Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 5 - Inner Product Spaces - 5.3 Orthonormal Bases: Gram-Schmidt Process - 5.3 Exercises - Page 257: 18

Answer

(a) the set is orthogonal. (b) The set $\{u_1,v_1\}$ $$u_1= (-\frac{2}{3}, \frac{1}{3},\frac{2}{3})$$ $$v_1= =(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}},0).$$ is orthonormal set.

Work Step by Step

Let $u=(-\frac{2}{15}, \frac{1}{15},\frac{2}{15})$ and $v=(\frac{1}{15}, \frac{2}{15},0)$, then we have (a) since $$u\cdot v=- \frac{2}{225}+ \frac{2}{225}=0,$$ then the set is orthogonal. (b) to normalize the set, we have $$u_1=\frac{1}{\|u\|}u=\frac{15}{3}(-\frac{2}{15}, \frac{1}{15},\frac{2}{15})=(-\frac{2}{3}, \frac{1}{3},\frac{2}{3})$$ $$v_1=\frac{1}{\|v\|}v=\frac{15}{\sqrt{5}}(\frac{1}{15}, \frac{2}{15},0)=(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}},0)$$ Now, the set $\{u_1,v_1\}$ is orthonormal set.
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