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 $m\times n$ matrix together with the Cartesian product $S_n\times S_m$ of all n-dimensional probability vectors $S_n$ and all m-dimensional probability vectors $S_m$; that is, $S_n:=\{x={(x_1,\dots ,x_n)}^T\in {\mathbb{R}}^n:x_i\geqq 0,{\sum }_{i=1}^n{x}_i=1\}$, where the symbol ^T denotes the transpose. A point $(\overline{x},\overline{y})\in S_n\times S_n$ is called an equilibrium point of a matrix game B if $x^{T}B\overline{y}\leqq {\overline{x}}^{T}B\overline{y}\leqq {\overline{x}}^{T}By$ for all $x,y\in S_n$ and $\overline{x}B\overline{y}=v$, where v is value of the game. If $n=m$ and B is skew symmetric, then we can check that $(\overline{x},\overline{y})\in S_n\times S_n$ is an equilibrium point of the game B if and only if $B\overline{x}\leqq 0$ and $B\overline{y}\leqq 0$. When B is an $n\times n$ skew symmetric matrix, $\overline{x}\in S_n$ is called a solution of the matrix game B if $B\overline{x}\leqq 0$ [1].

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

(LP)  Minimize $c^{T}x$  subject to $Ax\geqq b$, $x\geqq 0$,

(LD)  Maximize $b^{T}y$  subject to $A^{T}y\leqq c$, $y\geqq 0$,

where $c\in {\mathbb{R}}^n$, $x\in {\mathbb{R}}^n$, $b\in {\mathbb{R}}^m$, $y\in {\mathbb{R}}^m$, $A=[a_{ij}]$ is an $m\times n$ real matrix.

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

$$B=\left[\begin{array}{ccc}0& A^T& -c\\ -A& 0& b\\ c^T& -b^T& 0\end{array}\right].$$

Dantzig [1] gave the complete equivalence between the linear programming duality and the matrix game B. Many authors [2–5] 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 $x:=(x_1,\dots ,x_n)$ and $y:=(y_1,\dots ,y_n)$:

Consider the nonlinear programming problem (VOP):

where $X=\{x\in {\mathbb{R}}^n:g(x)\geqq b,x\geqq 0\}$, $f:{\mathbb{R}}^n\to {\mathbb{R}}^p$, $g:{\mathbb{R}}^n\to {\mathbb{R}}^m$ are continuously differentiable. The gradient $\mathrm{\nabla }f(x)$ is an $n\times p$ matrix, and $\mathrm{\nabla }g(x)$ is an $n\times m$ matrix.

Definition 2.1 [10]

A point $\overline{x}\in X$ is said to be an efficient solution for (VOP) if there exists no other feasible point $x\in X$ such that $(f_1(x),\dots ,f_p(x))\le (f_1(\overline{x}),\dots ,f_p(\overline{x}))$.

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

Definition 2.2 [6]

Let $B_i$, $i=1,\dots ,p$, be real $n\times n$ skew-symmetric matrices. A point $\overline{x}\in S_n$ is said to be a vector solution of the vector matrix game $B_i$, $i=1,\dots ,p$ if $({\overline{x}}^T{B}_{1}x,\dots ,{\overline{x}}^T{B}_{p}x)\nleqq ({\overline{x}}^T{B}_1\overline{x},\dots ,{\overline{x}}^T{B}_p\overline{x})\nleqq (x^T{B}_1\overline{x},\dots ,x^T{B}_p\overline{x})$ for any $x\in S_n$.

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

Lemma 2.1 [6]

Let $B_i$, $i=1,\dots ,p$, be an $n\times n$ skew symmetric matrix. Then $\overline{y}\in S_n$ is a vector solution of the vector matrix game $B_i$, $i=1,\dots ,p$, if and only if there exists $\xi \in {\stackrel{o}{S}}_p$ such that $({\sum }_{i=1}^p{\xi }_i{B}_i)\overline{y}\leqq 0$.

Remark 2.1 Let $B_i$, $i=1,\dots ,p$, be an $n\times n$ 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; $\overline{y}\in S_n$ is a vector solution of the vector matrix game $B_i$, $i=1,\dots ,p$, if and only if there exists $\xi \in {\stackrel{o}{S}}_p$ such that $\overline{y}\in F_{\xi }(\overline{y})$, where $F_{\xi }(x)=\{y\in S_n\mid y\in x-({\sum }_{i=1}^p{\xi }_i{B}_i)x-{\mathbb{R}}_{+}^n\}$.

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

Theorem 2.1 Let $B_i$, $i=1,\dots ,p$, be an $n\times n$ skew symmetric matrix. Then there exists a vector solution of the vector matrix game $B_i$, $i=1,\dots ,p$.

Proof Let $\xi \in {\stackrel{o}{S}}_p$. Define a multifunction $F_{\xi }:S_n\to S_n$ by, for any $x\in S_n$,

$$F_{\xi }(x)=\{y\in S_n|y\in x-\left(\sum _{i=1}^p{\xi }_i{B}_i\right)x-{\mathbb{R}}_{+}^n\}.$$

Then the multifunction $F_{\xi }$ is closed and hence upper semi-continuous, and so it follows from the well-known Kakutani fixed point theorem [11] that the multifunction $F_{\xi }$ has a fixed point. So, by Remark 2.1, there exists a vector solution of the vector matrix game $B_i$, $i=1,\dots ,p$. □

3 Equivalence relations

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

where $f:=(f_1,\dots ,f_p):{\mathbb{R}}^n\times {\mathbb{R}}^m\to {\mathbb{R}}^p$ are continuously differentiable.

Consider the vector matrix game defined by the following $(n+m+1)\times (n+m+1)$ skew symmetric matrix $B_i(x,y)$, $i=1,\dots ,p$, related to (SP) and (SD):

$$B_i(x,y)=\left[\begin{array}{ccc}0& -x{\mathrm{\nabla }}_y{f}_i{(x,y)}^T& -{\mathrm{\nabla }}_x{f}_i(x,y)\\ {\mathrm{\nabla }}_y{f}_i(x,y)x^T& 0& {\mathrm{\nabla }}_y{f}_i(x,y)\\ {\mathrm{\nabla }}_x{f}_i{(x,y)}^T& -{\mathrm{\nabla }}_y{f}_i{(x,y)}^T& 0\end{array}\right].$$

Now, we give equivalent relations between (SD) and the vector matrix game $B_i(x,y)$, $i=1,\dots ,p$.

Theorem 3.1 Let $(\overline{x},\overline{y},\overline{\xi })$ be feasible for (SP) and (SD), with ${\overline{y}}^T{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})={\overline{x}}^T{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)\times (\overline{x},\overline{y})=0$. Let $z^{\ast }=1/(1+{\sum }_i{\overline{x}}_i+{\sum }_j{\overline{y}}_j)$, $x^{\ast }=z^{\ast }\overline{x}$ and $y^{\ast }=z^{\ast }\overline{y}$. Then $(x^{\ast },y^{\ast },z^{\ast })$ is a vector solution of the vector matrix game $B_i(\overline{x},\overline{y})$, $i=1,\dots ,p$.

Proof Let $(\overline{x},\overline{y},\overline{\xi })$ be feasible for (SP) and (SD). Then the following holds:

(3.1)

(3.2)

(3.3)

(3.4)

Multiplying (3.3) by $\overline{x}\geqq 0$ gives $-\overline{x}{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f){(\overline{x},\overline{y})}^T\overline{y}=0$ and from (3.2),

$$-\overline{x}{\mathrm{\nabla }}_y\left({\overline{\xi }}^{T}f\right){(\overline{x},\overline{y})}^T\overline{y}-{\mathrm{\nabla }}_x\left({\overline{\xi }}^{T}f\right)(\overline{x},\overline{y})\leqq 0.$$

(3.5)

Multiplying (3.1) by ${\overline{x}}^T\overline{x}\geqq 0$, ${\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y}){\overline{x}}^T\overline{x}\leqq 0$. It implies that since ${\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})\leqq 0$,

$${\mathrm{\nabla }}_y\left({\overline{\xi }}^{T}f\right)(\overline{x},\overline{y}){\overline{x}}^T\overline{x}+{\mathrm{\nabla }}_y\left({\overline{\xi }}^{T}f\right)(\overline{x},\overline{y})\leqq 0.$$

(3.6)

From (3.3) we have

$${\mathrm{\nabla }}_x\left({\overline{\xi }}^{T}f\right){(\overline{x},\overline{y})}^T\overline{x}-{\mathrm{\nabla }}_y\left({\overline{\xi }}^{T}f\right){(\overline{x},\overline{y})}^T\overline{y}=0.$$

(3.7)

But $z^{\ast }>0$ 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:

$$(\sum _{i=1}^p{\overline{\xi }}_i{B}_i(\overline{x},\overline{y}))\left(\begin{array}{c}{x}^{\ast }\\ y^{\ast }\\ z^{\ast }\end{array}\right)\leqq 0.$$

By Lemma 2.1, $(x^{\ast },y^{\ast },z^{\ast })$ is a vector solution of the vector matrix game $B_i(\overline{x},\overline{y})$, $i=1,\dots ,p$. □

Theorem 3.2 Let $(x^{\ast },y^{\ast },z^{\ast })$ with $z^{\ast }>0$ be a vector solution of the vector matrix game $B_i(\overline{x},\overline{y})$, $i=1,\dots ,p$, where $\overline{x}=x^{\ast }/z^{\ast }$ and $\overline{y}=y^{\ast }/z^{\ast }$. Then there exists $\overline{\xi }\in {\stackrel{o}{S}}_p$ such that $(\overline{x},\overline{y},\overline{\xi })$ is feasible for (SP) and (SD), and ${\overline{y}}^T{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})={\overline{x}}^T{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})=0$. Moreover, if $f_i(\cdot ,y)$, $i=1,\dots ,p$, are convex for fixed y and $f_i(x,\cdot )$, $i=1,\dots ,p$, are concave for fixed x, then $(\overline{x},\overline{y})$ is efficient for (SP) with fixed $\overline{\xi }$ and $(\overline{x},\overline{y})$ is efficient for (SD) with fixed $\overline{\xi }$.

Proof Let $(x^{\ast },y^{\ast },z^{\ast })$ with $z^{\ast }>0$ be a vector solution of the vector matrix game $B_i(\overline{x},\overline{y})$, $i=1,\dots ,p$. Then by Lemma 2.1, there exists $\overline{\xi }\in {\stackrel{o}{S}}_p$ such that

$$(\sum _{i=1}^p{\overline{\xi }}_i{B}_i(\overline{x},\overline{y}))\left(\begin{array}{c}{x}^{\ast }\\ y^{\ast }\\ z^{\ast }\end{array}\right)\leqq 0.$$

Thus, we get

(3.11)

(3.12)

(3.13)

(3.14)

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

(3.15)

(3.16)

(3.17)

From (3.14),

$$\overline{x}\geqq 0,\overline{y}\geqq 0.$$

(3.18)

By (3.16), ${\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})({\overline{x}}^T\overline{x}+1)\leqq 0$. It implies that since ${\overline{x}}^T\overline{x}+1>0$,

$$-{\mathrm{\nabla }}_y\left({\overline{\xi }}^{T}f\right)(\overline{x},\overline{y})\geqq 0.$$

(3.19)

From (3.15), $-\overline{x}{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f){(\overline{x},\overline{y})}^T\overline{y}\leqq {\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})$. Using (3.18) and (3.19), we obtain $0\leqq -\overline{x}{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f){(\overline{x},\overline{y})}^T\overline{y}\leqq {\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})$. It implies that $-{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})\leqq 0$. From (3.17), ${\overline{x}}^T{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})\leqq {\overline{y}}^T{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})$. But since $\overline{x}\geqq 0$ and ${\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})\geqq 0$, ${\overline{x}}^T{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})\geqq 0$ and since $\overline{y}\geqq 0$ and ${\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})\leqq 0$, ${\overline{y}}^T{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})\leqq 0$. Then we have

$$0\leqq {\overline{x}}^T{\mathrm{\nabla }}_x\left({\overline{\xi }}^{T}f\right)(\overline{x},\overline{y})\leqq {\overline{y}}^T{\mathrm{\nabla }}_y\left({\overline{\xi }}^{T}f\right)(\overline{x},\overline{y})\leqq 0.$$

Hence, ${\overline{x}}^T{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})={\overline{y}}^T{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})$. Thus, $(\overline{x},\overline{y},\overline{\xi })$ is feasible for (SP) and (SD) with $f_i(\overline{x},\overline{y})-{\overline{y}}^T{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})=f_i(\overline{x},\overline{y})-{\overline{x}}^T{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})$, $i=1,\dots ,p$. Since $(\overline{x},\overline{y},\overline{\xi })$ is feasible for (SD), by weak duality in [9], $(f_1(x,y)-y^T{\mathrm{\nabla }}_y({\xi }^{T}f)(x,y),\dots ,f_p(x,y)-y^T{\mathrm{\nabla }}_y({\xi }^{T}f)(x,y))\nleqq (f_1(\overline{x},\overline{y})-{\overline{y}}^T{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y}),\dots ,f_p(\overline{x},\overline{y})-{\overline{y}}^T{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y}))$ and $(f_1(\overline{x},\overline{y})-{\overline{x}}^T{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y}),\dots ,f_p(\overline{x},\overline{y})-{\overline{x}}^T{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y}))\nleqq (f_1(u,v)-u^T{\mathrm{\nabla }}_u({\xi }^{T}f)(u,v),\dots ,f_p(u,v)-u^T{\mathrm{\nabla }}_u({\xi }^{T}f)(u,v))$ for any feasible $(u,v,\xi )$ of (SP) and (SD). Therefore, $(\overline{x},\overline{y})$ is efficient for (SP) with fixed $\overline{\xi }$ and $(\overline{x},\overline{y})$ is efficient for (SD) with fixed $\overline{\xi }$. □

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

Example 3.1 Let $f_1(x,y)=x^2-y^2$ and $f_2(x,y)=y-x$. 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 $B_i(x,y)$, $i=1,2$. Let

$$B_i(x,y)=\left(\begin{array}{ccc}0& -x{\mathrm{\nabla }}_y{f}_i{(x,y)}^T& -{\mathrm{\nabla }}_x{f}_i(x,y)\\ -{\mathrm{\nabla }}_y{f}_i(x,y)x^T& 0& {\mathrm{\nabla }}_y{f}_i(x,y)\\ {\mathrm{\nabla }}_x{f}_i{(x,y)}^T& -{\mathrm{\nabla }}_y{f}_i{(x,y)}^T& 0\end{array}\right).$$

Then

$$B_1(x,y)=\left(\begin{array}{ccc}0& 2xy& -2x\\ -2xy& 0& -2y\\ 2x& 2y& 0\end{array}\right)\text{and}{B}_2(x,y)=\left(\begin{array}{ccc}0& -x& 1\\ x& 0& 1\\ -1& -1& 0\end{array}\right).$$

Let $(x,y)\in {\mathbb{R}}^2$ and $(x^{\ast },y^{\ast },z^{\ast })\in S_3$ be a vector solution of the vector matrix game $B_i(x,y)$, $i=1,2$, if and only if there exist ${\xi }_1>0$, ${\xi }_2>0$, ${\xi }_1+{\xi }_2=1$ such that

$$\left({\xi }_1\right(\begin{array}{ccc}0& 2xy& -2x\\ -2xy& 0& -2y\\ 2x& 2y& 0\end{array})+{\xi }_2(\begin{array}{ccc}0& -x& 1\\ x& 0& 1\\ -1& -1& 0\end{array}\left)\right)\left(\begin{array}{c}{x}^{\ast }\\ y^{\ast }\\ z^{\ast }\end{array}\right)\leqq \left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$$

⇔ there exist ${\xi }_1>0$, ${\xi }_2>0$, ${\xi }_1+{\xi }_2=1$ such that

$$\left(\begin{array}{c}x(2y{\xi }_1-{\xi }_2)y^{\ast }-(2x{\xi }_1-{\xi }_2)z^{\ast }\\ -x(2y{\xi }_1-{\xi }_2)x^{\ast }-(2y{\xi }_1-{\xi }_2)z^{\ast }\\ (2x{\xi }_1-{\xi }_2)x^{\ast }+(2y{\xi }_1-{\xi }_2)y^{\ast }\end{array}\right)\leqq \left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$$

Thus, we determine the set of all the vector solutions of the vector matrix game $B_i(x,y)$, $i=1,2$.

• (I) the case that $x>0$:

• (a) $2x{\xi }_1-{\xi }_2>0$, $2y{\xi }_1-{\xi }_2>0$: $(x^{\ast },y^{\ast },z^{\ast })=(0,0,1)$.

• (b) $2x{\xi }_1-{\xi }_2>0$, $2y{\xi }_1-{\xi }_2=0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(0,\alpha ,1-\alpha )\mid 0\leqq \alpha \leqq 1\}$.

• (c) $2x{\xi }_1-{\xi }_2>0$, $2y{\xi }_1-{\xi }_2<0$: $(x^{\ast },y^{\ast },z^{\ast })=(0,1,0)$.

• (d) $2x{\xi }_1-{\xi }_2=0$, $2y{\xi }_1-{\xi }_2>0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(\alpha ,0,1-\alpha )\mid 0\leqq \alpha \leqq 1\}$.

• (e) $2x{\xi }_1-{\xi }_2=0$, $2y{\xi }_1-{\xi }_2=0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(x_1,x_2,x_3)\mid x_1\geqq 0,x_2\geqq 0,x_3\geqq 0,x_1+x_2+x_3=1\}$.

• (f) $2x{\xi }_1-{\xi }_2=0$, $2y{\xi }_1-{\xi }_2<0$: $(x^{\ast },y^{\ast },z^{\ast })=(0,1,0)$.

• (g) $2x{\xi }_1-{\xi }_2<0$, $2y{\xi }_1-{\xi }_2>0$: $(x^{\ast },y^{\ast },z^{\ast })=(1,0,0)$.

• (h) $2x{\xi }_1-{\xi }_2<0$, $2y{\xi }_1-{\xi }_2=0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(\alpha ,1-\alpha ,0)\mid 0\leqq \alpha \leqq 1\}$.

• (i) $2x{\xi }_1-{\xi }_2<0$, $2y{\xi }_1-{\xi }_2<0$: $(x^{\ast },y^{\ast },z^{\ast })=(0,1,0)$.

• (II) the case that $x=0$:

• (a) $2y{\xi }_1-{\xi }_2>0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(1-\alpha ,\alpha ,0)\mid \alpha \leqq \frac{{\xi }_2}{2y{\xi }_1},y>0,{\xi }_1>0,{\xi }_2>0,{\xi }_1+{\xi }_2=1\}$.

• (b) $2y{\xi }_1-{\xi }_2=0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(\alpha ,1-\alpha ,0)\mid 0\leqq \alpha \leqq 1\}$.

• (c) $2y{\xi }_1-{\xi }_2<0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(\alpha ,1-\alpha ,0)\mid 0\leqq \alpha \leqq 1\}$.

• (III) the case that $x<0$:

• (a) $2y{\xi }_1-{\xi }_2>0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(\frac{2y{\xi }_1-{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2},-\frac{2x{\xi }_1-{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2},-\frac{2xy{\xi }_1-x{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2})$: $2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2>0,2y{\xi }_1-{\xi }_2>0,2x{\xi }_1-{\xi }_2<0,{\xi }_1>0,{\xi }_2>0,{\xi }_1+{\xi }_2=1\}$.

• (b) $2y{\xi }_1-{\xi }_2=0$: $(x^{\ast },y^{\ast },z^{\ast }):\{(\alpha ,1-\alpha ,0)\mid 0\leqq \alpha \leqq 1\}$.

• (c) $2y{\xi }_1-{\xi }_2<0$: $(x^{\ast },y^{\ast },z^{\ast })=(1,0,0)$.

Let $(x,y)\in {\mathbb{R}}^2$ and $S_{(x,y)}$ be the set of vector solutions of the vector matrix game $B_i(x,y)$, $i=1,2$. From (I), (II) and (III),

$$\begin{array}{rcl}\bigcup _{(x,y)\in {\mathbb{R}}^2}S(x,y)& =& \{(\alpha ,1-\alpha ,0)\mid 0\leqq \alpha \leqq 1\}\cup \{(0,\alpha ,1-\alpha )\mid 0\leqq \alpha \leqq 1\}\\ \cup \{(\alpha ,0,1-\alpha )\mid 0\leqq \alpha \leqq 1\}\\ \cup \{(\alpha ,\beta ,\gamma )\mid \alpha \geqq 0,\beta \geqq 0,\gamma \geqq 0,\alpha +\beta +\gamma =1\}\\ \cup \left\{\right(\frac{2y{\xi }_1-{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2},-\frac{2x{\xi }_1-{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2},\\ -\frac{2xy{\xi }_1-x{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2}\left)\right|x<0,2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2>0,\\ 2y{\xi }_1-{\xi }_2>0,2x{\xi }_1-{\xi }_2<0,{\xi }_1>0,{\xi }_2>0,{\xi }_1+{\xi }_2=1\}.\end{array}$$

Let $(\overline{x},\overline{y},\overline{\xi })$ be feasible for (SP) and (SD) with $\overline{y}{\mathrm{\nabla }}_y({\overline{\xi }}^{T}f)(\overline{x},\overline{y})=\overline{x}{\mathrm{\nabla }}_x({\overline{\xi }}^{T}f)(\overline{x},\overline{y})=0$. We can easily check that

Thus,

Therefore, Theorem 3.1 holds.

Let $(x,y)\in {\mathbb{R}}^2$ and $S_{(x,y)}$ be the set of vector solutions of the vector matrix game $B_i(x,y)$, $i=1,2$. Then

$$\begin{array}{rcl}\bigcup _{(x,y)\in {\mathbb{R}}^2}{S}_{(x,y)}& =& \{(\alpha ,1-\alpha ,0)\mid 0\leqq \alpha \leqq 1\}\cup \{(0,\alpha ,1-\alpha )\mid 0\leqq \alpha \leqq 1\}\\ \cup \{(\alpha ,0,1-\alpha )\mid 0\leqq \alpha \leqq 1\}\\ \cup \{(\alpha ,\beta ,\gamma )\mid \alpha \geqq 0,\beta \geqq 0,\gamma \geqq 0,\alpha +\beta +\gamma =1\}\\ \cup \left\{\right(\frac{2y{\xi }_1-{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2},-\frac{2x{\xi }_1-{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2},\\ -\frac{2xy{\xi }_1-x{\xi }_2}{2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2}\left)\right|x<0,2y{\xi }_1-2x{\xi }_1-2xy{\xi }_1+x{\xi }_2>0,\\ 2y{\xi }_1-{\xi }_2>0,2x{\xi }_1-{\xi }_2<0,{\xi }_1>0,{\xi }_2>0,{\xi }_1+{\xi }_2=1\}.\end{array}$$

So,

$$\left\{(\frac{x^{\ast }}{z^{\ast }},\frac{y^{\ast }}{z^{\ast }})\right|z^{\ast }>0\text{ and }(x^{\ast },y^{\ast },z^{\ast })\in S_{(\frac{x^{\ast }}{z^{\ast }},\frac{y^{\ast }}{z^{\ast }})}\}=\left\{(\frac{{\xi }_2}{2{\xi }_1},\frac{{\xi }_2}{2{\xi }_1})\right|{\xi }_1>0,{\xi }_2>0,{\xi }_1+{\xi }_2=1\}.$$

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 $\{(\frac{{\xi }_2}{2{\xi }_1},\frac{{\xi }_2}{2{\xi }_1},{\xi }_1,{\xi }_2)\mid {\xi }_1>0,{\xi }_2>0,{\xi }_1+{\xi }_2=1\}\subset F\cap G$ and $(\frac{{\xi }_2}{2{\xi }_1}){\mathrm{\nabla }}_y({\xi }^{T}f)(\frac{{\xi }_2}{2{\xi }_1},\frac{{\xi }_2}{2{\xi }_1})=(\frac{{\xi }_2}{2{\xi }_1}){\mathrm{\nabla }}_x({\xi }^{T}f)(\frac{{\xi }_2}{2{\xi }_1},\frac{{\xi }_2}{2{\xi }_1})=0$. Therefore, Theorem 3.2 holds.