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Unformatted text preview: UC Berkeley Department of Electrical Engineering and Computer Science EECS 227A Nonlinear and Convex Optimization Solutions 1 Fall 2009 The first part of this problem set provides some practice on the mathematical prerequisites for this course (vector calculus, elementary analysis, and linear algebra) and will calibrate your degree of preparation for the course. Reading: Boyd and Vandenberghe, § 3.1, 3.2 and Appendix A for background material. Solution 1.1 Consider the eigenvector decomposition Q = n summationdisplay i =1 λ i u i u T i . First, suppose that Q is positive semidefinite. In this case, we must have u T k Qu k ≥ 0 for all k = 1 ,...,n . But since the eigenvectors { u i } are all orthonormal, we have u T k Qu k = λ k ≥ for all k = 1 ,...,n . In the other direction, suppose that all eigenvalues of Q are nonnegative (i.e. λ k ≥ 0 for all k = 1 ,...,n ). Then for any x ∈ R n , we have x T Qx = n summationdisplay i =1 λ i ( x T u i ) 2 ≥ so Q is positive semidefinite. Solution 1.2 By decomposing x = bracketleftbigg x A x C bracketrightbigg , we have x T Mx = x T A Ax A + 2 x T A Bx C + x T C Cx C (a) ⇒ (b) Let x C = 0, since x T Mx > , ∀ x negationslash = 0 we also have x T A Ax A > , ∀ x A negationslash = 0, thus A is positive definite (so it is invertible). Let x A = − A − 1 Bx C , then x T Mx = x T C ( C − B T A − 1 B ) x C . Again, since x T Mx > , ∀ x negationslash = 0 we also have x T C ( C − B T A − 1 B ) x C > , ∀ x C negationslash = 0, thus C − B T A − 1 B is positive definite (b) ⇒ (a) min x negationslash =0 x T Mx = min braceleftbigg min x C negationslash =0 min x A x T A Ax A + 2 x T A Bx C + x T C Cx C , min x C =0 min x A negationslash =0 x T A Ax A + 2 x T A Bx C + x T C Cx C bracerightbigg = min braceleftbigg min x C negationslash =0 min x A x T A Ax A + 2 x T A Bx C + x T C Cx C , min x A negationslash =0 x T A Ax A bracerightbigg Since A is positive definite, assuming that x C is fixed, x T A Ax A + 2 x T A Bx C + x T C Cx C achieves minimum at x A = − A − 1 Bx C , thus min x negationslash =0 x T Mx = min braceleftbigg min x C negationslash...
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This note was uploaded on 09/03/2011 for the course EE 227A taught by Professor Staff during the Spring '11 term at Berkeley.
 Spring '11
 Staff
 Electrical Engineering

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