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# Lecture-11 - Lecture-11 Theorems 5.3 and 5.2 Algorithms 5.1...

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1 Lecture-11 Theorems 5.3 and 5.2 Algorithms 5.1, 5.2 Theorem 5.3 1. The directions are indeed conjugate. 2. Therefore, the algorithm terminates in n steps (from Theorem 5.1). 3. The residuals are mutually orthogonal. 4. Each direction p k and r k is contained in Krylov subspace of r 0 degree k .

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2 Theorem 5.3 (1) 1 , , 0 for 0 - = = k i r r i T k K Suppose that the k th iteration generated by the conjugate gradient method is not the solution point x * . The following four properties hold: Therefore, the sequence { x k } converges to x * in at most n steps. { } { } (2) , , , span , , , span 0 0 0 1 0 r A Ar r r r r k k K K = (4) 1 , , 0 for 0 - = = k i Ap p i T k K { } { } (3) , , , span , , , span 0 0 0 1 0 r A Ar r p p p k k K K = Proof { } { } { } { } (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 • Use induction on (2) and (3) • First prove (2) • Then prove (3) using (2) • Prove (4) by induction using (3) and Theorem 5.2 •Prove (1) using (4) and Theorem 5.2
3 Proof Induction: k=0 { } { } { } { } (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 (2) And (3) { } { } (2) span span 0 0 r r = 0 0 r p - = { } { } (3) span span 0 0 r p = Proof { } { } 0 0 0 0 0 0 , , , span , , , , span r A Ar r p r A Ar r r k k k k K K { } 0 1 0 2 0 , , , span r A r A Ar Ap k k + K { } 0 1 0 0 1 , , , span r A Ar r r k k + + K Assume (2) and (3) are true for k , prove for k+ 1 To prove (2), by induction : By multiplying with A k k k k Ap r r a + = + 1 By combining this with induction hypothesis on (2) { } { } { } { } (4) 1 , , 0 for 0 (3) , , , span , , , span (2) , , , span , , , span

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Lecture-11 - Lecture-11 Theorems 5.3 and 5.2 Algorithms 5.1...

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