Differential Equations Solutions 14

Differential Equations Solutions 14 - ± ≤ j − 1, since...

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24 Chapter 5. Solutions: Matrix Factorizations Alternatively, using Gram-Schmidt orthogonalization, r 11 = p 3 2 +3 2 =3 2 , q 1 = 1 3 2 · 3 3 ¸ . Then r 12 = q T 1 · 3 1 ¸ =4 / 2 , b q 2 = · 3 1 ¸ 4 / 2 q 1 , and r 22 = the norm of this vector = 2, so q 2 = b q 2 / 2. If we complete the arithmetic, we get the same QR as above, up to choice of signs for columns of Q and rows of R . CHALLENGE 5.9. Note that after we Fnish the iteration i =1 ,w ehav e q new k +1 = q old k +1 r 1 ,k +1 q 1 ,so q 1 q new k +1 = q 1 q old k +1 r 1 ,k +1 q 1 q 1 =0 by the deFnition of r 1 ,k +1 and the fact that q 1 q 1 =1 . Assume that after we Fnish iteration i = j 1, for a given value of k ,w ehav e q ± q k +1 =0fo r ± j 1and q j q ± =0fo r j<± k . After we Fnish iteration i = j for that value of k ,wehave q j q new k +1 = 0 by the same argument we used above, and we also have that
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Unformatted text preview: ± ≤ j − 1, since all we have done to q k +1 is to add a multiple of q j to it, and q j is orthogonal to q ± . Thus, after iteration j , q ∗ ± q new k +1 = 0 for ± ≤ j , and the induction is complete when j = k and k = n − 1. CHALLENGE 5.10. (a) We verify that Q is unitary by showing that its conjugate transpose is its inverse: Q ∗ Q = ( I − 2 uu ∗ )( I − 2 uu ∗ ) = I − 4 uu ∗ + 4 uu ∗ uu ∗ = I , since u ∗ u = 1. ±or the second part, we compute v ∗ z = ( z ∗ − α ∗ e T 1 ) z = z ∗ z − α ∗ z 1 = z ∗ z − e − iθ ± z ± e iθ ζ = z ∗ z − ± z ± ζ,...
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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