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: 12

Answer

We can write the given vector form as $\left[\begin{array}{l} s+3t\\ s-t\\ 2s-t\\ 4t \end{array}\right]=s\left[\begin{array}{l} 1\\ 1\\ 2\\ 0 \end{array}\right]+t\left[\begin{array}{l} 3\\ -1\\ -1\\ 4 \end{array}\right], s,t\in \mathbb{R}$ $ u=\left[\begin{array}{l} 1\\ 1\\ 2\\ 0 \end{array}\right], v=\left[\begin{array}{l} 3\\ -1\\ -1\\ 4 \end{array}\right],\quad$ Since $u,v\in \mathbb{R}^{4}$ and $W=$Span$\{u,v\}$, by Theorem 1, $W$ is a subspace of $\mathbb{R}^{4}$.

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. ------------- We can write the given vector form as $\left[\begin{array}{l} s+3t\\ s-t\\ 2s-t\\ 4t \end{array}\right]=s\left[\begin{array}{l} 1\\ 1\\ 2\\ 0 \end{array}\right]+t\left[\begin{array}{l} 3\\ -1\\ -1\\ 4 \end{array}\right], s,t\in \mathbb{R}$ $ u=\left[\begin{array}{l} 1\\ 1\\ 2\\ 0 \end{array}\right], v=\left[\begin{array}{l} 3\\ -1\\ -1\\ 4 \end{array}\right],\quad$ Since $u,v\in \mathbb{R}^{4}$ and $W=$Span$\{u,v\}$, by Theorem 1, $W$ is a subspace of $\mathbb{R}^{4}$.
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