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: 10

Answer

$\left[\begin{array}{l} 2t\\ 0\\ -t \end{array}\right]=t\left[\begin{array}{l} 2\\ 0\\ -1 \end{array}\right]$ $v=\left[\begin{array}{l} 2\\ 0\\ 1 \end{array}\right]\in \mathbb{R}^{3}$ and $H=$Span$\{v\}$, so by Theorem 1, $H$ is a subspace of $\mathbb{R}^{3}$.

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. ------------- $\left[\begin{array}{l} 2t\\ 0\\ -t \end{array}\right]=t\left[\begin{array}{l} 2\\ 0\\ -1 \end{array}\right]$ $v=\left[\begin{array}{l} 2\\ 0\\ 1 \end{array}\right]\in \mathbb{R}^{3}$ and $H=$Span$\{v\}$, so by Theorem 1, $H$ is a subspace of $\mathbb{R}^{3}$.
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