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 - 5.1 Length and Dot Product in Rn - 5.1 Exercises - Page 236: 68

Answer

(a) ${u} \cdot {v}=5$. (b) ${v} \cdot {v}=45$. (c) $\|u\|^{2}=50$. (d) $({u} \cdot {v})v=\left[\begin{array}{l}{0}\\ {10} \\ {25}\\ {20}\end{array}\right]$. (e) ${u} \cdot {(5v)}={u}^{T} {(5v)}=25$.

Work Step by Step

Given ${u}=\left[\begin{array}{l}{4} \\ {0}\\ {-3} \\ {5}\end{array}\right]$ and ${v}=\left[\begin{array}{l}{0}\\ {2} \\ {5} \\ {4} \end{array}\right]$, then we have (a) ${u} \cdot {v}={u}^{T} {v}=[4 \ \ 0 \ \ -3 \ \ 5 ]\left[\begin{array}{1} {0}\\ {2} \\ {5} \\ {4}\end{array}\right]=[(4)(0)+(0)(2)+(-3)(5)+(5)(4)]=5$. (b) ${v} \cdot {v}={v}^{T} {v}=[0 \ \ 2 \ \ 5 \ \ 4 ]\left[\begin{array}{1} {0}\\ {2} \\ {5} \\ {4}\end{array}\right]=[(0)(0)+(2)(2)+(5)(5)+(4)(4)]=45$. (c) $\|u\|^{2}={u} \cdot {u}={u}^{T} {u}=[4 \ \ 0 \ \ -3 \ \ 5 ]\left[\begin{array}{l}{4} \\ {0}\\ {-3} \\ {5}\end{array}\right]=[(4)(4)+(0)(0)+(-3)(-3)+(5)(5)]=50$. (d) $({u} \cdot {v})v=({u}^{T} {v}) v=5\left[\begin{array}{l}{0}\\ {2} \\ {5} \\ {4} \end{array}\right]=\left[\begin{array}{l}{0}\\ {10} \\ {25}\\ {20}\end{array}\right]$. (e) ${u} \cdot {(5v)}={u}^{T} {(5v)}=[4 \ \ 0 \ \ -3 \ \ 5 ]\left[\begin{array}{1} {0}\\ {10} \\ {25} \\ {20}\end{array}\right]=[(4)(0)+(0)(10)+(-3)(25)+(5)(20)]=25$.
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