A Note on Gordan's Theorem

On a class of theorems equivalent to Farkas's lemma

“…This approach offers a way to understand these theorems better, and it offers a way to bypass the difficulty of attacking a problem directly. For example, if one knows that Farkas's lemma is equivalent to Gordan's theorem, then in order to prove Farkas's lemma, it suffices to prove Gordan's theorem ( [12]). Similarly, by going through Sep I ⇒ Mangasarian ⇒ Theorem 17 ⇒ Broyden (see Section 3 (d)), this gives a new proof of Broyden's theorem.…”

“…(a) That 1, 2, 3, 4, and 5 are equivalent was proven in[12], where 1 and 4 are essentially tautology. (b) 5 ⇒ 7 ⇒ 6 ⇒ 5.…”

“…We may decompose V asV = W + V 0 (eg. see Lemma 4.1 of[12]), where W is a maximal linear subspace and V 0 is pointed (the case when V = W is easy, so we assume here that V 0 = {0}). By the propertiesof V ∩ N = {0},it is straightforward to check that N + (−1)V 0 is pointed and [N + (−1)V 0 ] ∩ W = {0}.…”