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 $\mathbb{R}^{3}$.
