20095ee103_1_hw4_sols

20095ee103_1_hw4_sols - L. Vandenberghe 10/22/09 EE103...

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Unformatted text preview: L. Vandenberghe 10/22/09 EE103 Homework 4 solutions 1. Exercise 6.4 (b). This matrix is positive definite. Applying the definition, we have to show that x T Ax = x T ( I uu T ) x > for all nonzero x . We can simplify x T Ax as x T Ax = x T x x T uu T x = bardbl x bardbl 2 ( u T x ) 2 . It follows from the Cauchy-Schwarz inequality that | x T u | bardbl x bardblbardbl u bardbl < bardbl x bardbl if bardbl u bardbl < 1 and bardbl x bardbl negationslash = 0. Therefore bardbl x bardbl 2 ( u T x ) 2 > for all nonzero x . 2. Exercise 6.12. If we work out the product in the factorization we get K = bracketleftBigg L 11 L T 11 L 11 L T 21 L 21 L T 11 L 21 L T 21 L 22 L T 22 bracketrightBigg . This gives three conditions A = L 11 L T 11 , B = L 11 L T 21 , C + L 21 L T 21 = L 22 L T 22 . This shows that L 11 is the Cholesky factor of A , and L 22 is the Cholesky factor of C + L 21 L T 21 . This last matrix is positive definite because x T ( C + L 21 L T 21 ) x = x T Cx + x T L 21 L T 21 x = x T Cx + bardbl L T 21 x bardbl 2 2 > for all nonzero x if C is positive definite....
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This note was uploaded on 09/28/2010 for the course EE EE 103 taught by Professor Jacobsen during the Fall '09 term at UCLA.

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20095ee103_1_hw4_sols - L. Vandenberghe 10/22/09 EE103...

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