Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 4 - Vector Spaces - 4.1 Exercises - Page 198: 17

Answer

$u=\left[\begin{array}{l} 1\\ 0\\ -1\\ 0 \end{array}\right], v=\left[\begin{array}{l} -1\\ 1\\ 0\\ 1 \end{array}\right],w=\left[\begin{array}{l} 0\\ -1\\ 1\\ 0 \end{array}\right].$ $W=$Span$ \{u,v,w\}$,

Work Step by Step

See Theorem 1, p.196 If $v_{1},v_{2}...,v_{p}$ are in a vector space $V$, then Span $\{v_{1},v_{2}...,v_{p}\}$ is a subspace of V. ------------ $\left[\begin{array}{l} a-b\\ b-c\\ c-a\\ b \end{array}\right]=a\left[\begin{array}{l} 1\\ 0\\ -1\\ 0 \end{array}\right]+b\left[\begin{array}{l} -1\\ 1\\ 0\\ 1 \end{array}\right]+c\left[\begin{array}{l} 0\\ -1\\ 1\\ 0 \end{array}\right]$ $u=\left[\begin{array}{l} 1\\ 0\\ -1\\ 0 \end{array}\right], v=\left[\begin{array}{l} -1\\ 1\\ 0\\ 1 \end{array}\right],w=\left[\begin{array}{l} 0\\ -1\\ 1\\ 0 \end{array}\right].$ $u,v,w\in \mathbb{R}^{3}$,and $W=$Span$ \{u,v,w\}$, so , by Theorem 1, $W$ is a subspace of $\mathbb{R}^{3}.$
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