Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.3 - The Dot Product - Exercises 12.3 - Page 713: 26


Proof given below.

Work Step by Step

By Property 4 of the Dot Product (boxed on p. 709) ${\bf u}\cdot{\bf u}=|{\bf u}|^{2}\geq 0$, which proves the first condition. Proof of the second condition: If ${\bf u}={\bf 0}$, then $|{\bf u}|=0$, and ${\bf u}\cdot{\bf u}=0.$ Thus, ${\bf u}={\bf 0}\Rightarrow{\bf u}\cdot{\bf u}=0$ If ${\bf u}\cdot{\bf u}$=$0$, if we assume that ${\bf u}\neq{\bf 0}$, it would follow that $|{\bf u}|^{2}\neq 0,$ which is a contradiction to ${\bf u}\cdot{\bf u}=|{\bf u}|^{2}$=$0$ Thus, ${\bf u}\cdot{\bf u}=0\Rightarrow{\bf u}={\bf 0}$. which proves that dot multiplication is positive definite.
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