## Linear Algebra and Its Applications (5th Edition)

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 $\mathbb{R}^{4}$.
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 $\mathbb{R}^{4}$.