Answer
$u=\left[\begin{array}{l}
5\\
1\\
0
\end{array}\right], v=\left[\begin{array}{l}
2\\
0\\
1
\end{array}\right]$
Reason: Theorem 1.
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 as
$\left[\begin{array}{l}
5b+2c\\
b\\
c
\end{array}\right]=b\left[\begin{array}{l}
5\\
1\\
0
\end{array}\right]+c\left[\begin{array}{l}
2\\
0\\
1
\end{array}\right], b,c\in \mathbb{R}$
$ u=\left[\begin{array}{l}
5\\
1\\
0
\end{array}\right], v=\left[\begin{array}{l}
2\\
0\\
1
\end{array}\right],\quad$
Since $ u,v\in \mathbb{R}^{3}$ and $H=$Span$\{u,v\}$,
by Theorem 1,
$H$ is a subspace of $ \mathbb{R}^{3}$.
