Find two vectors for basis R4 I have to find two vectors for a basis in $\mathbb{R}^{4}$and that basis needs to contain $ v = \begin{pmatrix} 1\\2\\-1\\0\end{pmatrix},u =\begin{pmatrix} 1\\0\\1\\3\end{pmatrix}$. Then the unitvectors $e2$ and $e4$ are two vectors for a basis together with $v$ and $u$, because these two vectors don't have a $1$ at the second and fourth spot, right?
Edit: $ e2 = \begin{pmatrix} 0 \\ 1 \\0 \\ 0\end{pmatrix}$ and $e4 = \begin{pmatrix} 0 \\ 0 \\0 \\ 1\end{pmatrix}$
 A: If I get your question correct, you want to extend the reduced basis $(\mathcal{B_2})$ of  $\mathbb{R}^{4}$. If yes, then $e_2$ and $e_4$ are not the vectors, with which $\mathcal{B_2}$ can be extended to $\mathcal{B_4}$.
If you are looking forward to extend $\mathcal{B_2}$, you may try to do the following: create a matrix $A_{4\times6}(\mathbb{R})$, such that its columns for sure span the whole $\mathbb{R^4}$:
$$ A = \left(\begin{array}{cc|cccc}
1 & 1 & 1 & 0 & 0 & 0\\
2 & 0 & 0 & 1 & 0 & 0\\
-1 & 1 & 0 & 0& 1 & 0\\
 0 & 3 & 0 & 0& 0 & 1\\
\end{array}\right) \xrightarrow[]{\text{RREF}} \left(\begin{array}{cc|cccc}
1 & 0 & 0 & 0 & -1 & \frac{1}{3}\\
0 & 1 & 0 & 0 & 0 & \frac{1}{3}\\
0 & 0 & 1 & 0 & 1 & -\frac{2}{3}\\
0 & 0 & 0 & 1 & 2 & -\frac{2}{3}\\
\end{array}\right)$$
Now, you have to pick the leading colums to the right of the dash in $RREF(A)$, i.e. the $3^{rd}$ and the $4^{th}$ columns of $RREF(A)$, and take corresponding columns in $A$ as the ones, which extend the basis $\mathcal{B_2}$ to $\mathcal{B_4}$ of $\mathbb{R^{4}}$. That is $\mathcal{B_4} = \mathcal{B_2} \cup \left\{e_1 = \begin{pmatrix}
1\\
0\\
0\\
0\\
\end{pmatrix}, e_2 = \begin{pmatrix}
0\\
1\\
0\\
0\\
\end{pmatrix}\right\}.$
If you need something else, then sorry for wasting your time :)
