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