# WUSTL ESE 415 - tutorial5 (4 pages)

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## tutorial5

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## tutorial5

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4
School:
Washington University in St. Louis
Course:
Ese 415 - Optimization

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ESE 415 U S Kamilov Tutorial session 2 23 2018 TA Hesam Mazidi Problem 1 Show that the Vandermonde matrix 4 x A x3 x2 x3 x2 x x2 x 1 1 is positive semidefinite PSD but not positive definite PD Solution 1 We note that A is PSD if and only if for any y y1 y2 y3 T we have y T Ay 0 Further A is PD if and only if for any y y1 y2 y3 T 6 0 0 0 T we have y T Ay 0 Notice that for any y y1 y2 y3 T y T Ay y1 x2 y2 x y3 2 2 Thus it immediately follows that A is PSD Now assume y y1 y2 y3 T 6 0 0 0 T In this case we can always find a vector y1 y2 y3 T such that y1 x2 y2 x y3 0 In particular 2 x x2 satisfies this condition concluding that A is not PD as desired Problem2 In each of the following problems fully justify your answer using optimality conditions i Show that f x y x2 4 2 y 2 has two global minima and one stationary point which is neither a local maximum nor a local minimum ii Find all local minima and all local maxima of f x y sin x sin y sin x y within the set 0 x 2 0 y 2 Page 1 of 4 ESE 415 U S Kamilov Tutorial session 2 23 2018 TA Hesam Mazidi Solution 2 i Note that f is a differentiable function with domain R2 This means that any stationary point x y T local maximum global maximum etc must satisfy f x y T 0 That is 2 x 2 4 2x 4 x 3 4x f 0 3 2y 2y The stationary points are easily obtained form eq 3 as x y T 0 0 T 2 0 T or 2 0 T Since f is twice differentiable we can invoke sufficient second order optimality conditions In particular denote H x y T as the Hessian of f obtained at x y T Then if H x y T is positive definite it follows that x y T is a local maximum if H x y T is negative definite x y T is a local minimum H x y T 4 3x 2 4 0 0 2 4 We therefore have the following characterizations 2 0 T is a local minimum since H 2 0 T is positive definite 2 0 T is a local minimum since H 2 0 T is positive definite Note that H 0 0 T is indefinite that is it is neither positive semidefinite nor negative semidefinite By invoking the necessary second order optimality conditions

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