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assingnment3-sol

# assingnment3-sol - Assignment 3 Solution November 4 2008...

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Assignment 3: Solution November 4, 2008 Question 1 1. By Gauss elimination £ -→ u 1 -→ u 2 -→ u 3 / = 1 1 2 1 3 1 2 1 3 1 4 1 3 1 4 1 5 -→ 1 1 2 1 3 0 1 12 1 12 0 0 1 180 hence -→ u 1 , -→ u 2 , -→ u 3 are linear independent. Then we do Gram-Schmidt process to get an orthonormal basis: -→ e 1 = -→ u 1 || -→ u 1 || = 6 7 , 3 7 , 2 7 T = [0 . 8571 , 0 . 4286 , 0 . 2857] T ; -→ v 2 = -→ u 2 - -→ u 2 · -→ e 1 -→ e 1 = - 5 98 , 17 294 , 13 196 T -→ e 2 = -→ v 2 || -→ v 2 || = - 30 3577 , 34 3577 , 39 3577 T = [ - 0 . 5016 , 0 . 5685 , 0 . 6521] T ; -→ v 3 = -→ u 3 - -→ u 3 · -→ e 1 -→ e 1 - -→ u 3 · -→ e 2 -→ e 2 = 1 2190 , - 1 365 , 1 365 T -→ e 3 = -→ v 3 || -→ v 3 || = 1 73 , - 6 73 , 6 73 T = [0 . 1170 , - 0 . 7022 , 0 . 7022] T . Each step, we get a non-zero vector, which can be normalized. This is since the original vector -→ u 1 , -→ u 2 , -→ u 3 are linear independent. 1

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2. Assume -→ u 1 , -→ u 2 , . . . , -→ u k are linear independent. For i = 1 , -→ e 1 = -→ u 1 || -→ u 1 || is an orthonormal basis. Assume for i = n (1 n k - 1), we get an orthonormal basis -→ e 1 , -→ e 2 , . . . , -→ e n from -→ u 1 , -→ u 2 , . . . , -→ u n . For i = n + 1, define --→ v n +1 = --→ u n +1 - --→ u n +1 · -→ e 1 -→ e 1 - . . . - --→ u n +1 · -→ e n -→ e n . First we show that --→ v n +1 6 = -→ 0 . Otherwise --→ u n +1 = --→ u n +1 · -→ e 1 -→ e 1 + . . . + --→ u n +1 · -→ e n -→ e n span { -→ e 1 , -→ e 2 , . . . , -→ e n } = span { -→ u 1 , -→ u 2 , . . . , -→ u n } which contradicts that -→ u 1 , -→ u 2 , . . . , --→
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assingnment3-sol - Assignment 3 Solution November 4 2008...

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