hw1sol_fa09

hw1sol_fa09 - UC Berkeley Department of Electrical...

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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 pre-requisites 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 non-negative (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.

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hw1sol_fa09 - UC Berkeley Department of Electrical...

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