Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 5 - Inner Product Spaces - 5.1 Definition of an Inner Product Space - Problems - Page 352: 37

Answer

See below

Work Step by Step

Let $w,v \in V$ $||v+w||^2\\ =(\sqrt )^2\\ =\\ =+\\ =\bar{}+\bar{}\\ =\bar{}+\bar{}+\bar{}+\bar{}\\ =||v||^2+++||w||^2$ We have $z=a+bi\\ \rightarrow z+\bar z=a+bi+a-bi=2a=2Re(z)$ Hence, $+\bar{}=2Re()$ Consequently, $||v+w||^2=||v||^2+2Re()+||w||^2$
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