20105ee103_1_hw1_sols

20105ee103_1_hw1_sols - L. Vandenberghe 9/30/10 EE103...

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Unformatted text preview: L. Vandenberghe 9/30/10 EE103 Homework 1 solutions 1. Exercise 1.2. (a) Not linear or affine. Choose x = (1 , 0), y = (0 , 1), α = β = 1 / 2. Then f ( αx + βy ) = 0 negationslash = αf ( x ) + βf ( y ) = 1 . (b) Linear. We have f ( x ) = a T x for a = ( − 1 , , . . . , , 1). (c) Affine. Working out the squares gives f ( x ) = ( x − c ) T ( x − c ) − ( x − d ) T ( x − d ) = ( x T x − c T x − x T c + c T c ) − ( x T x − d T x − x T d + d T d ) = x T x − 2 c T x + c T c − x T x + 2 d T x − d T d = 2( d − c ) T x + bardbl c bardbl 2 − bardbl d bardbl 2 . Hence, f can be expressed as f ( x ) = a T x + b with a = 2( d − c ) and b = bardbl c bardbl 2 −bardbl d bardbl 2 . The function is linear if bardbl c bardbl = bardbl d bardbl . 2. Exercise 1.9. bardbl a + b bardbl 2 = ( a + b ) T ( a + b ) = a T a + b T a + a T b + b T b = bardbl a bardbl 2 + 2 a T b + bardbl b bardbl 2 ≥ bardbl a bardbl 2 − 2 bardbl a bardblbardbl b bardbl + bardbl b bardbl 2 = ( bardbl a bardbl − bardbl b bardbl ) 2 . The fourth line follows from the Cauchy-Schwarz inequality. 3. Exercise 1.10. (a) We can determine t by setting the derivative of f ( t ) = bardbl tx − y bardbl 2 = t 2 bardbl x bardbl 2 − 2 tx T y + bardbl y bardbl 2 equal to zero. This gives the condition 2 t bardbl x bardbl 2 − 2 x T y = 0. The solution is= 0....
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This note was uploaded on 02/03/2011 for the course EE 103 taught by Professor Vandenberghe,lieven during the Spring '08 term at UCLA.

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20105ee103_1_hw1_sols - L. Vandenberghe 9/30/10 EE103...

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