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

More documents

Recommendations

Info

1346 Cherng-tiao Perng It is easy to see that there do not exist ū i ≽ 0 such that ∑ ū i ᾱ i =¯γ. (3) i
A note on D. Bartl’s algebraic proof of Farkas’s lemma 1347 Of course this is essentially what the author of [2] had, namely ᾱ i = α i − ιλ i α m and ¯γ = γ − ιvα m , where λ i = αm −1 (x 1 )α i (x 1 ) and v = αm −1 (x 1 )γ(x 1 ). 2. We note here that the result of the generalized Farkas’s Lemma above can be reduced to the case when W is a finite dimensional vector space easily. By linear algebra, W = W ′ ⊕ ker A is a direct sum of W ′ and the kernel of A, where W ′ ≃ imA is a finite dimensional vector space. Assume that condition (B) in the Farkas’s Lemma is false, we proceed to establish (A). If γ |ker A is not trivial, then there exists x ∈ ker A such that γ(x) ≺ 0 and α i (x) =0, ∀i. Therefore, we may well assume that γ |ker A is trivial, and restrict everything, including the map A, toW ′ , but this is just the case for finite dimensional space. Closely related to Farkas’s Lemma is Gordan’s Theorem. In the following formulation, we have to assume that V is nontrivial, i.e. V ≠ {0}. Gordan’s Theorem 3.2 Let A : W → F m be a linear map and let o : W → V be the zero map, where V ≠ {0}. Then either (A) there exists an x ∈ W such that α i (x) > 0, 1 ≤ i ≤ m, or (B) there exist nonnegative vectors u 1 , ···,u m ∈ V , not all zero, such that u 1 α 1 + ···+ u m α m = o. The alternatives (A) and (B) exclude each other. Proof. Clearly (A) and (B) cannot be both true. Assume that (B) is false, we will prove by induction that there exists x such that α i (x) > 0 for i ≤ m. For m = 1, that there does not exist u 1 ≻ 0 such that u 1 α 1 = o implies that α 1 is not the zero map, therefore there exists x ∈ W such that α 1 (x) > 0. Assume the result is true for the case m − 1, we will show the case m. By assumption, there exists x 1 ∈ W such that α i (x 1 ) > 0 for all i 0, we would be finished. It remains to handle the cases α m (x 1 )=0 and α m (x 1 ) < 0. Case 1: α m (x 1 )=0. By the same reasoning as in the base case, α m cannot be the zero map, hence there exists y ∈ W such that α m (y) > 0. Now if
Page 1 and 2: International Mathematical Forum, V
Page 3: A note on D. Bartl’s algebraic pr
Page 7: A note on D. Bartl’s algebraic pr

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

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?