20095ee103_1_hw5_sols

20095ee103_1_hw5_sols - L. Vandenberghe 11/05/09 EE103...

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Unformatted text preview: L. Vandenberghe 11/05/09 EE103 Homework 5 solutions 1. Exercise 8.1. We can note that A = PD where P is a permutation matrix, and D is diagonal: A = 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 10 2 10 3 10 4 10 . (a) We have bardbl A bardbl = max x negationslash =0 radicalBig (10 2 x 1 ) 2 + (10 3 x 2 ) 2 + ( 10 4 x 3 ) 2 + ( 10 x 4 ) 2 radicalBig x 2 1 + x 2 2 + x 2 3 + x 2 4 = 10 4 . The maximum in the definition of A is obtained by choosing x 1 = 0 , x 2 = 0 , x 3 = 1 , x 4 = 0 . (The argument is the same as in the derivation of the norm of a diagonal matrix.) (b) We can find the inverse by solving AX = I , column by column. We can also use the expression A = PD and write A 1 = ( PD ) 1 = D 1 P 1 = D 1 P T . Therefore A 1 = 10 2 10 3 10 4 10 1 0 0 0 1 0 0 1 0 1 0 0 0 0 1 0 0 = 10 2 10 3 10 4 10 1 . (c) As in part (a), we have bardbl A bardbl = max x negationslash =0 radicalBig ( 10 4 x 1 ) 2 + ( 10 1 x 2 ) 2 + (10 3 x 3 ) 2 + (10 2 x 4 ) 2 radicalBig x 2 1 + x 2 2 + x 2 3 + x 2 4 = 10 3 . 1 (d) ( A ) = bardbl A bardblbardbl A 1 bardbl = 10 7 . 2. Exercise 8.3. bardbl A bardbl bardbl Ax bardbl / bardbl x bardbl for all x negationslash = 0. Therefore bardbl A bardbl max { bardbl Ax (1) bardbl bardbl x (1) bardbl , bardbl Ax (2) bardbl bardbl x (2) bardbl , bardbl Ax (3) bardbl bardbl x (3) bardbl , bardbl Ax (4) bardbl bardbl x (4) bardbl...
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This note was uploaded on 05/10/2010 for the course EE EE103 taught by Professor Jacobson during the Spring '09 term at UCLA.

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20095ee103_1_hw5_sols - L. Vandenberghe 11/05/09 EE103...

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