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: 28

Answer

$B'=\{ (\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}),(-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})\}$ is orthonormal basis for $R^2$.

Work Step by Step

Let $B=\{ v_1=(1,2),v_2=(-1,0) \}$. Applying the Gram-Schmidt orthonormalization process produces $$w_1=v_1=(1,2)$$ $$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(-1,0)+\frac{1}{5}(1,2)=(-\frac{4}{5},\frac{2}{5}).$$ Normalizing $w_1$and $w_2$ produces the orthonormal set $$u_1=\frac{w_1}{\|w_1\|}=(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$$ $$u_2=\frac{w_2}{\|w_2\|}=\frac{\sqrt{5}}{2}(-\frac{4}{5},\frac{2}{5})= (-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}).$$ Hence, $B'=\{ (\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}),(-\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}})\}$ is orthonormal basis for $R^2$.
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