Lecture-12 - Lecture-12 Theorems 5.3 and 5.2 Algorithms...

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1 Lecture-12 Theorems 5.3 and 5.2 Algorithms 5.1, 5.2 Proof k i Ap p Ap r Ap p i T k k i T k i T k , , 1 , 0 for 1 1 1 K = + - = + + + b k i p r i T k , , 0 for 0 1 K = = + Now Conjugacy (4): ; 1 1 1 k k k k p r p + + + + - = b ; 1 1 k T k k T k k Ap p Ap r + + b Due to this the right side becomes Zero for i=k By induction hypothesis on (4) the vectors are conjugate up to p k By Theorem 5.2 { } { } { } { } (4) 1 , , 0 for 0 (3) , , , span , , , span (2) , , , span , , , span (1) 1 , , 0 for 0 0 0 0 1 0 0 0 0 1 0 - = = = = - = = k i Ap p r A Ar r p p p r A Ar r r r r k i r r i T k k k k k i T k K K K K K K By definition : By definition : Therefore (F) (4) Holds for k= 1 (4) 0 0 1 = Ap p T Assume true for k , prove true for k +1
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2 Proof { } { } { } 1 1 0 0 1 0 2 0 0 0 0 , , , span , , , span , , , span + + = i i i i p p p r A r A Ar r A Ar r A Ap K K K 1 , , 0 for 0 1 - = = + k i Ap r i T k K { } { } { } { } (4) 1 , , 0 for 0 (3) , , , span , , , span (2) , , , span , , , span (1) 1 , , 0 for 0 0 0 0 1 0 0 0 0 1 0 - = = = = - = = k i Ap p r A Ar r p p p r A Ar r r r r k i r r i T k k k k k i T k K K K K K K By applying (3) k i p r i T k , , 0 for 0 1 K = = + k i Ap p Ap r Ap p i T k k i T k i T k , , 1 , 0 for 1 1 1 K = + - = + + + b So the first term vanishes in (F) . Due to induction hypothesis on (4) the second term vanishes as well. Hence QED (4). So the direction set generated by CG method is indeed a conjugate direction set. According to Theorem 5.1 the algorithm terminates in at most
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This note was uploaded on 06/12/2011 for the course COT 6505 taught by Professor Shah during the Spring '07 term at University of Central Florida.

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Lecture-12 - Lecture-12 Theorems 5.3 and 5.2 Algorithms...

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