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: 9

Answer

$v=\left[\begin{array}{l} 1\\ 3\\ 2 \end{array}\right] $. Reason: Theorem 1.

Work Step by Step

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. ------------- $\left[\begin{array}{l} s\\ 3s\\ 2s \end{array}\right]=s\left[\begin{array}{l} 1\\ 3\\ 2 \end{array}\right]$ $v=\left[\begin{array}{l} 1\\ 3\\ 2 \end{array}\right]\in \mathbb{R}^{3}$ and $H=$Span$\{\left[\begin{array}{l} 1\\ 3\\ 2 \end{array}\right]\}$, so by theorem 1, $H$ is a subspace of $ \mathbb{R}^{3}$.
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