farkas_handout - Farkas Lemma Rudi Pendavingh Eindhoven...

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Farkas’ Lemma Rudi Pendavingh Eindhoven Technical University Optimization in R n , lecture 2 Rudi Pendavingh (TUE) Farkas’ Lemma ORN2 1 / 14
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Today’s Lecture Theorem (Farkas’ Lemma, 1894) Let A be an m × n matrix, b R n . Then either: 1 There is an x R n such that Ax b; or 2 There is a y R m such that y 0 , yA = 0 and yb < 0 . 1 Farkas’ Lemma is at the core of linear optimization. 2 There are extremely efficient proofs of Farkas’ Lemma. We’ll do a slow, geometrical proof. 3 Farkas’ Lemma is augmented by Carath´ eodory’s theorem . Rudi Pendavingh (TUE) Farkas’ Lemma ORN2 2 / 14
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Convex sets Definition Let x , y R n . The line segment between x and y is [ x , y ] := { λ x + (1 - λ ) y | λ [0 , 1] } . Definition A set C R n is convex if [ x , y ] C for all x , y C . Example any affine space { x R n | Ax = b } is convex any halfspace { x R n | ax β } is convex any polyhedron { x R n | Ax b } is convex the unit ball { x R n | k x k ≤ 1 } is convex any ellipsoid { Ax + b | k x k ≤ 1 } is convex Rudi Pendavingh (TUE) Farkas’ Lemma ORN2 3 / 14
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Separation Definition A set H R n is a hyperplane if H = { x R n | dx = δ } =: H d for some nonzero rowvector d R n and δ R . Definition Let X , Y R n . Then H d separates X from Y if dx < δ for all x X , dy > δ for all y Y . Theorem (’The Separation Theorem’)
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farkas_handout - Farkas Lemma Rudi Pendavingh Eindhoven...

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