Answer
$u=\left[\begin{array}{l}
4\\
0\\
1\\
-2
\end{array}\right], v=\left[\begin{array}{l}
3\\
0\\
1\\
0
\end{array}\right],w=\left[\begin{array}{l}
0\\
0\\
1\\
1
\end{array}\right]$
$W=$Span$ \{u,v,w\}$
Work Step by Step
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.
------------
$\left[\begin{array}{l}
4a+3b\\
0\\
a+b+c\\
c-2a
\end{array}\right]=a\left[\begin{array}{l}
4\\
0\\
1\\
-2
\end{array}\right]+b\left[\begin{array}{l}
3\\
0\\
1\\
0
\end{array}\right]+c\left[\begin{array}{l}
0\\
0\\
1\\
1
\end{array}\right]$
$u=\left[\begin{array}{l}
4\\
0\\
1\\
-2
\end{array}\right], v=\left[\begin{array}{l}
3\\
0\\
1\\
0
\end{array}\right],w=\left[\begin{array}{l}
0\\
0\\
1\\
1
\end{array}\right].$
$u,v,w\in \mathbb{R}^{4}$,and $W=$Span$ \{u,v,w\}$,
so , by Theorem 1,
$W$ is a subspace of $\mathbb{R}^{4}.$
