Is $U \subseteq V$ for given $U$ and $V$? How can I decide if it is true, that $U \subseteq V$ for given 
$$
U = \operatorname{span}\{(−1,−1,−2,2,1),(3,2,3,1,−3),(1,0,−1,5,−1),(1,−2,−3,3,1)\}\\
V=\{(x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5\mid x_1 + 3x_2 − x_3 + 2x_5= 0\}
$$
And also, how to expand basis of a space $U$ to get a basis of a space $V$?
 A: Note that $V$ is a subspace, so $U \subset V$ iff the four elements defining $U$ are in $V$. 
Note that $V = \{ x \mid \langle n, x \rangle = 0 \}$,
where $n=(1,3,-1,0,2)^T$. Hence
$\dim V = 4$, and you know $\dim U = 3$. You need to find one more vector.
One way would be to find an element in the kernel of
$\begin{bmatrix}
& & u_1^T \\& &  u_2^T \\& &  u_3^T \\& &  u_4^T \\
1 & 3 & -1 & 0 & 2
\end{bmatrix}$, where $u_k$ are the vectors
constituting $V$.
Another approach would be to use Gram Schmidt on the vectors
$n, u_1,u_2,u_3,u_4, e_1,e_2,...$, ignoring vectors are are in the span of previous ones.
Hint: If you apply the process outlined in the previous step and divide the
vector corresponding to $e_1$ by its smallest modulus element, you obtain the
following vector:

 $(1,-3,8,3,1)^T$

A: Since $U$ and $V$ are vector spaces, $U$ is a subset of $V$ if each of the specified vectors that together span $U$ is in $V$.  And all you need to do in this case to see whether one of those is a member of $V$ is substituted its coordinates for $x_1,x_2,x_3,x_4,x_5$ and see whether they satisfy the equation that defines $V$.
The dimension of the domain of the map $(x_1,x_2,x_3,x_4,x_5)\mapsto x_1 + 3x_2 − x_3 + 2x_5$ is $5$ and the dimension of the image is $1$, so the dimension of the kernel must be $5-1=4$, and the kernel is $V$.  So $\dim V=4$.  Since there are four vectors in $U$, they will span $V$ if they are linearly independent, but not if they're not.
