Ee1034deS10

Ee1034deS10 - Recursive LU Factorization A LU A is nxn a11...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Recursive LU Factorization ? ,x A LU A is n n 11 11 1 11 10 0 tt n nn ad ug fL eD U     1 1 x (1 ) , ) x 1 , ) x ) ) x ) t n di s n ei s n D i n L and U are n n 1 ua g d f e  11 11 11 ,, a ? 11 11 1 11 1 1 t n a D ed L U  11 1 0 n a e L e DU 11 ? 1 1 t n a or De d L U  EE103 Lec 4DE (SEJ) 1 An LU Factorization Code EE103 Lec 4DE (SEJ) 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
PD Matrices: Choleski Factorization (Examples of when PD Matrices Occur?) Let A be an m x n matrix, mn and assume the columns of A are linearly independent. Then A A T is symmetric and positive definite. Ex 1: 1: ( ) TT T AA 2: 0. 0 0 def T Let x Then x A Ax y y yA x   EE103 Lec 4DE (SEJ) 3 Example: Unconstrained Minimization min ( , , ) min ( ) n f x x x f x 1 2 Ex 2: Unconstrained Minimization ,,, x x x x n 12 EE103 Lec 4DE (SEJ) 4
Background image of page 2
i i f Example: Unconstrained Minimization min ( ,, ) min ( ) ,,, xx x n x n f x x x f x 12 First Order Necessary Condition for a Local Minimum f x () 0 Second Order Sufficient Condition for a Local Minimum () f Hx i s P D 22 2 ff f        2 1 1 n f x f x x x      2 2 f fx  EE103 Lec 4DE (SEJ) 5 nn n x x x  Least Squares via Choleski, 2 nd degree approximating polynomial 1.0 1 y x 21 0 1.0 42 1 5 aa a a   1.5 2 32 . 0 1.0 4 55 . 0 0 0 0 0 .5 93 2 . 0 16 4 1.0 25 5 5.0 a a a 65 . 5 0 36 6 5.5 a EE103 Lec 4DE (SEJ) 6
Background image of page 3

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

View Full DocumentRight Arrow Icon
Least Squares via Choleski, Interpolating Polynomial >> xd=linspace(1,6); 1.0 1 1.5 2 y x   >> yd=polyval(a,xd); >> plot(x,y,'ro',xd,yd) 32 . 0 1.0 4 55 . 0 65 5  8 65.5 54 3 2 1 0 3 2 1 0 543 2 1 0 1.0 32 16 8 4 2 1.5 243 81 27 9 3 2.0 aa a a a a a a a a aaa a a a  5 6 7 3 2 1 0 2 1 0 3 2 1 0 1024 256 64 16 4 1.0 3125 625 125 25 5 5.0 7776 1296 216 36 6 5.5 a a a a a a a a a a a 3 4 >> X=[x.^5 x.^4 x.^3 x.^2 x ones(length(x), 1)] 1 2 3 4 5 6 0 1 2 EE103 Lec 4DE (SEJ) 7 Min Norm, using other Norms 1 1 1. n i i zz   2 2 1 1 2. 3. max n t i i i i z z  n ,, n 1 1 2 2 1 1. ; 2. ; 3m a x ; i i n t i i z z black z z b l u e d o t t e d   1, , 3. max i in EE103 Lec 4DE (SEJ) 8
Background image of page 4
Choleski Factorization 11 11 21 31 41 ? 21 22 22 32 42 000 11 2 3 00 0 15 4 7 ll l l l lll   Example of “recursive Choleski” 31 32 33 33 43 41 42 43 44 44 0 241 41 1 371 11 5 A llll l     11 11 21 31 41 ? 21 22 22 32 42 1123 00 0 154 7 0 0 0 1 0 l l l         31 32 33 33 43 41 42 43 44 44 3 7 11 15 0 0 0 l     t
Background image of page 5

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

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 13

Ee1034deS10 - Recursive LU Factorization A LU A is nxn a11...

This preview shows document pages 1 - 6. Sign up to view the full document.

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