A Note on D. Bartl's Algebraic Proof of Farkas's Lemma 1 Introduction

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1344 Cherng-tiao Perng the sense that the current proof is more conceptual, and the algebra involved is totally down-to-earth. In passing, we also prove Gordan’s Theorem, using similar technique. <strong>Note</strong> that in the classical setting, with the ground field replaced by any linearly ordered field, we know that many key results such as Farkas’s Lemma, variant of Farkas’s Lemma ([1]), some versions of separating hyperplane theorems (pertaining to linear polyhedral cones), Gale’s Theorem (see [7]), and Gordan’s Theorem ([6]) (this list can be enlarged to include most of the theorems of alternatives discussed in [7]) are actually equivalent (see [8]). This observation opens the door for future investigation of generalizations for the related theorems. 2 Notations F : a linearly ordered field, not necessarily commutative. W : a right vector space over F , whose dimension may be finite or infinite. V : a right vector space over F . α i : linear functional on W , i.e. α i : W → F is a linear map. We endow the space of linear functionals with a left vector space structure. We make this choice so that the expression aα i (xb) =aα i (x)b makes natural sense, where a, b ∈ F, and x ∈ W. γ: a linear transformation from W to V , compatible with the vector space structures of W and V , namely γ(xb) =γ(x)b, for all x ∈ W and all b ∈ F , etc. The symbol A stands for the linear transformation A : W → F m . A linear transformation A is equivalent to the information of m linear functionals, α i : W → F ,1≤ i ≤ m, where α i (x) represents the i-th component of the column vector Ax with x ∈ W. We denote the ordering of F and V by the symbols “≤” and “≼”, respectively. For the ordering, the following five statements must hold true for u, v ∈ V and any μ ∈ F so that V is a linearly ordered vector space over the linearly ordered field F : i) u ≼ v if and only if u − v ≼ 0 ii) u ≽ 0oru ≼ 0 iii) if u ≽ 0 and u ≼ 0, then u =0 iv) if u ≽ 0 and v ≽ 0, then u + v ≽ 0 v) if μ ≥ 0 and u ≽ 0, then uμ ≽ 0
A note on D. Bartl’s algebraic proof of Farkas’s lemma 1345 3 Main Results We will prove the following generalized form of Farkas’s Lemma, which is Lemma 2 of [2]. Farkas’s Lemma 3.1 Let A : W → F m and γ : W → V be linear mappings. Then either (A) there exists an x ∈ W such that α 1 (x) ≥ 0, ···,α m (x) ≥ 0, and γ(x) ≺ 0, or (B) there exist nonnegative vectors u 1 , ···,u m ∈ V such that γ = u 1 α 1 + ···+ u m α m . The alternatives (A) and (B) exclude each other. Proof. First of all, for any m, it is clear that (A) and (B) cannot be both true, for this would imply that γ(x) =u 1 α 1 (x)+···+ u m α m (x) ≽ 0 contradicting the condition γ(x) ≺ 0. For the remaining part, we will assume that (B) is false, and proceed to establish (A). We prove this by induction on m. For m = 1, that (B) is false implies that γ ≠ 0, otherwise letting u 1 = 0 would give γ = u 1 α 1 , contrary to the assumption. Therefore there exists x 1 ∈ W such that γ(x 1 ) ≺ 0. Now if α 1 (x 1 ) ≥ 0, we would be finished. Therefore assume that α 1 (x 1 ) < 0. We define ¯γ by ¯γ = γ −γ(x 1 )α 1 (x 1 ) −1 α 1 . Necessarily ¯γ ≠0, otherwise γ would satisfy (B). Since ¯γ ≠0, there exists ¯x ∈ W such that ¯γ(¯x) ≺ 0. Letting now x = ¯x − x 1 α 1 (x 1 ) −1 α 1 (¯x), it is straightforward to check that γ(x) =¯γ(¯x) ≺ 0, and α 1 (x) =0, hence the case of m = 1 is proven. We assume that the case m − 1 ≥ 1 is true, and prove for the case m. Assume (B) is false, i.e. there do not exist u i ≽ 0 such that ∑ m i=1 u iα i = γ, which also implies that there do not exist u i ≽ 0 such that ∑ i
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A Note on D. Bartl's Algebraic Proof of Farkas's Lemma 1 Introduction

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