MATH 2107assign2sol_09

MATH 2107assign2sol_09 - MATH 2107 LINEAR ALGEBRA II...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 2107 LINEAR ALGEBRA II ASSIGNMENT 2 DUE: December 4 at the beginning of the tutorial 1. Let = 2 2 2 2 5 4 2 4 5 A (a) Show that the matrix A is positive definite. ANS: 0 5 ) det( ) 1 ( > = A , 0 9 5 4 4 5 ) det( ) 2 ( > = = A , 0 10 2 . 2 ) 4 .( 4 6 . 5 2 2 5 4 ) 2 ( 2 2 2 4 ) 4 ( 2 2 2 5 5 ) det( ) det( ) 3 ( > = + = + = = A A Therefore the matrix A is positive definite. (b) Find the Cholesky factorization of the matrix A . 1 45 50 5 2 5 9 9 2 5 6 5 2 5 2 5 9 5 2 5 4 0 0 0 2 4 5 0 0 2 4 5 2 2 2 2 5 4 2 4 5 2 3 ' 3 1 1 3 ' 3 1 2 ' 2 U A R R R R R R R R R = = = + = + = = 45 50 5 3 2 5 3 5 2 5 4 0 0 0 5 U Check: A U U t = = = 2 2 2 2 5 4 2 4 5 0 0 0 5 0 0 0 5 45 50 5 3 2 5 3 5 2 5 4 45 50 5 3 2 5 2 5 3 5 4 (c) Orthogonally diagonalize the matrix A . λ = = = 2 4 2 2 9 4 0 0 1 2 2 2 2 5 4 0 1 1 2 2 2 2 5 4 2 4 5 ) det( I A ) 10 )( 1 )(( 1 ( ) 10 11 )( 1 ( ) 8 ) 2 )( 9 )(( 1 ( 2 = + = =
Background image of page 1

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

View Full DocumentRight Arrow Icon
So, the eigenvalues of A are 1(with multiplicity 2) and 10. I. Eigenvectors for 1 = λ : = 0 0 0 0 0 0 1 2 2 ~ 1 2 2 2 4 4 2 4 4 . 1 I A which implies that 3 2 , x x are free variables. So, + = + = = 2 0 1 2 / 0 1 1 2 / ) 2 ( 0 ) ( 3 2 3 2 3 2 x x x x x x X X I A , 3 2 , x x Therefore the eigenvectors corresponding to 1 = are = 0 1 1 1 v and = 2 0 1 2 v II. Eigenvectors for 10 = : = 0 0 0 2 1 0 4 1 1 ~ 8 2 2 2 5 4 2 4 5 . 10 I A which implies that 3 x is free variable. So, ⎡− = ⎡− = = 1 2 2 2 2 0 ) 10 ( 3 3 3 3 x x x x X X I A 3 x Therefore the eigenvectors corresponding to 10 = is ⎡− = 1 2 2 3 v The eigenvectors 2 1 , v v are orthogonal to 3 v , but 2 1 , v v are not orthogonal to each other. By Gram-Schmidt algorithm , = = = 2 2 / 1 2 / 1 0 1 1 2 1 2 0 1 . . 1 1 1 1 2 2 2 v v v v v v z 2 1 , z v are orthogonal to each other.
Background image of page 2
By normalizing the vectors 3 2 1 , , v z v , we get = 0 2 / 1 2 / 1 1 u , = 2 3 / 4 2 3 / 1 2 3 / 1 2 u , ⎡− = 3 / 1 3 / 2 3 / 2 3 u Let = 3 / 1 2 3 / 4 0 3 / 2 2 3 / 1 2 / 1 3 / 2 2 3 / 1 2 / 1 P and
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 10

MATH 2107assign2sol_09 - MATH 2107 LINEAR ALGEBRA II...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online