Find a basis for $\mathbb{R}^4$ containing two vectors that form a basis for the null space of the given matrix A \begin{bmatrix}
  1 & 2 & 1 & 1 \\
  2 & 4 & 3 & 5\\
 \end{bmatrix}
When I reduce this matrix to reduced row echelon form and set Ax=0, I get
\begin{bmatrix}
  1 & 2 & 0 & -2 \\
  0 & 0 & 1 & 3 \\
 \end{bmatrix}
I've found the vectors forming the basis of the nullspace of A: $(2,−1,0,0),(2,0,−3,1)$ What would the other two vectors be? $(1,0,0,0)$ and $(0,1,0,0)$?
 A: I have no idea from where from where the $-2$ and the $2$ in the reduced echolon form comes. 
The idea is at first find two linear independet vectors which are in the null space. (for example by the reduced echolon form), afterward add some vectors for a basis.
A: Here is a general procedure for completing the basis of a space. Suppose you already have a few vectors that are the basis of a subspace (it does not matter what the subspace is). Let these vectors be 
$$v_1, v_2, v_3, \cdots v_k$$
Now pick any basis for the space (columns/rows of identity) is an obvious choice. Let these vectors be
$$u_1, u_2, u_3 \cdots u_n$$
Now run the gram-schmdit process on the $k+n$ vectors ordered as $v$'s followed by $u$'s, i.e.
$$
[v_1, v_2 \cdots v_k, u_1, u_2, \cdots u_n] \rightarrow [w_1, w_2, \cdots w_k, w_{k+1}, \cdots w_n] ~~\hbox{Using Gram-Schmidt}
$$
Now $w_1, w_2 \cdots w_k$ is an orthonormal basis for the original space spanned by $v_1, \cdots v_k$ and the rest complete the space (actually forms a basis for the orthogonal complemnt
