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 - 5.2 Inner Product Spaces - 5.2 Exercises - Page 247: 93


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Work Step by Step

Since for any $w\in W$, we have $$\langle 0,w\rangle =0$$ then $0\in W^\perp$ and $W$ in nonempty set. Now, let $v_1,v_2\in W^\perp$, then for any $w\in W$, we get \begin{aligned} \langle v_1+v_2, w\rangle &=\langle v_1,w\rangle +\langle v_2,w\rangle=0 \end{aligned} hence, $v_1+v_2\in W^\perp$ which means that $W^\perp$ is closed under addition. For any $v\in W^\perp$, $w\in W$ and $k\in R$, we have $$\langle kv, w\rangle =k\langle v, w\rangle=0.$$ Hence, $W^\perp$ is closed scalar multiplication. Consequently, $W^\perp$ is a subspace of $W$.
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