{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

20095ee103_1_hw4_sols

# 20095ee103_1_hw4_sols - L Vandenberghe EE103 Homework 4...

This preview shows pages 1–2. Sign up to view the full content.

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 > 0 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 bardbl bardbl 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 > 0 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 > 0 for all nonzero x if C is positive definite. The factorization can be computed as follows.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

20095ee103_1_hw4_sols - L Vandenberghe EE103 Homework 4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online