hw1sol - EE364a Winter 2007-08 Prof S Boyd EE364a Homework...

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Unformatted text preview: EE364a, Winter 2007-08 Prof. S. Boyd EE364a Homework 1 solutions 2.1 Let C ⊆ R n be a convex set, with x 1 , . . . , x k ∈ C , and let θ 1 , . . . , θ k ∈ R satisfy θ i ≥ 0, θ 1 + ··· + θ k = 1. Show that θ 1 x 1 + ··· + θ k x k ∈ C . (The definition of convexity is that this holds for k = 2; you must show it for arbitrary k .) Hint. Use induction on k . Solution. This is readily shown by induction from the definition of convex set. We illustrate the idea for k = 3, leaving the general case to the reader. Suppose that x 1 , x 2 , x 3 ∈ C , and θ 1 + θ 2 + θ 3 = 1 with θ 1 , θ 2 , θ 3 ≥ 0. We will show that y = θ 1 x 1 + θ 2 x 2 + θ 3 x 3 ∈ C . At least one of the θ i is not equal to one; without loss of generality we can assume that θ 1 negationslash = 1. Then we can write y = θ 1 x 1 + (1 − θ 1 )( μ 2 x 2 + μ 3 x 3 ) where μ 2 = θ 2 / (1 − θ 1 ) and μ 2 = θ 3 / (1 − θ 1 ). Note that μ 2 , μ 3 ≥ 0 and μ 1 + μ 2 = θ 2 + θ 3 1 − θ 1 = 1 − θ 1 1 − θ 1 = 1 . Since C is convex and x 2 , x 3 ∈ C , we conclude that μ 2 x 2 + μ 3 x 3 ∈ C . Since this point and x 1 are in C , y ∈ C . 2.2 Show that a set is convex if and only if its intersection with any line is convex. Show that a set is affine if and only if its intersection with any line is affine. Solution. We prove the first part. The intersection of two convex sets is convex. Therefore if S is a convex set, the intersection of S with a line is convex. Conversely, suppose the intersection of S with any line is convex. Take any two distinct points x 1 and x 2 ∈ S . The intersection of S with the line through x 1 and x 2 is convex. Therefore convex combinations of x 1 and x 2 belong to the intersection, hence also to S . 2.5 What is the distance between two parallel hyperplanes { x ∈ R n | a T x = b 1 } and { x ∈ R n | a T x = b 2 } ? Solution. The distance between the two hyperplanes is | b 1 − b 2 | / bardbl a bardbl 2 . To see this, consider the construction in the figure below. 1 a x 1 = ( b 1 / bardbl a bardbl 2 ) a x 2 = ( b 2 / bardbl a bardbl 2 ) a a T x = b 2 a T x = b 1 The distance between the two hyperplanes is also the distance between the two points x 1 and x 2 where the hyperplane intersects the line through the origin and parallel to the normal vector a . These points are given by x 1 = ( b 1 / bardbl a bardbl 2 2 ) a, x 2 = ( b 2 / bardbl a bardbl 2 2 ) a, and the distance is bardbl x 1 − x 2 bardbl 2 = | b 1 − b 2 | / bardbl a bardbl 2 . 2.7 Voronoi description of halfspace. Let a and b be distinct points in R n . Show that the set of all points that are closer (in Euclidean norm) to a than b , i.e. , { x | bardbl x − a bardbl 2 ≤ bardbl x − b bardbl 2 } , is a halfspace. Describe it explicitly as an inequality of the form c T x ≤ d ....
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hw1sol - EE364a Winter 2007-08 Prof S Boyd EE364a Homework...

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