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: 31

Answer

$B'=\{(\frac{1}{3},-\frac{2}{3},\frac{2}{3}),(\frac{2}{3},\frac{2}{3},\frac{1}{3}),(\frac{2}{3},-\frac{1}{3},-\frac{2}{3})\}$ is the orthonormal basis for $R^3$.

Work Step by Step

Let $B=\{ v_1=(1,-2,2),v_2=(2,2,1), v_3=(2,-1,-2) \}$. Applying the Gram-Schmidt orthonormalization process produces $$w_1=v_1=(1,-2,2)$$ $$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(2,2,1)-0(1,-2,2)=(2,2,1)$$ $$w_3=v_3- \frac{v_3\cdot w_1}{w_1\cdot w_1}w_1- \frac{v_3\cdot w_2}{w_2\cdot w_2}w_2\\=(2,-1,-2)-0(1,-2,2)-0(2,2,1)=(2,-1,-2).$$ Normalizing $w_1$, $w_2$ and $w_3$ produces the orthonormal set $$u_1=\frac{w_1}{\|w_1\|}=\frac{1}{3}(1,-2,2)=(\frac{1}{3},-\frac{2}{3},\frac{2}{3})$$ $$u_2=\frac{w_2}{\|w_2\|}=\frac{1}{3}(2,2,1)=(\frac{2}{3},\frac{2}{3},\frac{1}{3})$$ $$u_3=\frac{w_3}{\|w_3\|}=\frac{1}{3}(2,-1,-2)=(\frac{2}{3},-\frac{1}{3},-\frac{2}{3})$$ Hence, $B'=\{(\frac{1}{3},-\frac{2}{3},\frac{2}{3}),(\frac{2}{3},\frac{2}{3},\frac{1}{3}),(\frac{2}{3},-\frac{1}{3},-\frac{2}{3})\}$ is the orthonormal basis for $R^3$.
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