How do I complete a set of three vectors in $\mathbb{R}^4$ to a basis of that space? Given vectors $$w_1 (0,-1,2,1),\quad w_2 (1,0,2,1),\quad w_4 (2,1,1,0),$$
how do I find another vector $v$ such that $\{w_1, w_2, w_4, v\}$ is linearly independent?
My approach is to write them all in matrix form and then try and row-reduce them, but if one row is just $a,b,c,d$ and others with numbers, how will I be able to show it is a basis? Or am I missing a point here?
Can I just take the 3 given vectors and row-reduce them? 
 A: If you place the three given vectors as rows of a matrix and do row-reduction, the set of vectors spanned by the rows after reduction will be the same as the set spanned before reduction (assuming you do it right). But AFTER reduction, it should be easier to find a vector not in the span of your rows. For instance, if you ended up with 
$$
\begin{bmatrix}
1 & 0 & 3 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix},
$$
then the vector $[0, 0, 1, 0]$ would not be in the row span. 
A: A set of $n$ vectors in $\mathbb{R}^n$ is linearly independent iff the $n \times n$ matrix produced by adjoining them is nonsingular, which means we need to find a vector $(a, b, c, d)^T$ such that
$$\det\begin{pmatrix}0 & 1 & 2 & a \\ -1 & 0 & 1 & b \\ 2 & 2 & 1 & c \\ 1 & 1 & 0 & d \end{pmatrix} \neq 0.$$
You could expand this, which would take some work, or you can observe that, e.g., the vectors in $\mathbb{R}^3$ produced by projecting the $w_i$ onto the first three components are themselves linearly independent, so if we set $a = b = c = 0$, we have
$$\det\begin{pmatrix}0 & 1 & 2 & 0 \\ -1 & 0 & 1 & 0 \\ 2 & 2 & 1 & 0 \\ 1 & 1 & 0 & d \end{pmatrix} \neq 0 = -(d) \det \begin{pmatrix}0 & 1 & 2 \\ -1 & 0 & 1 \\ 2 & 2 & 1 \end{pmatrix} \neq 0,$$ provided $d \neq 0$. So, any vector $(0, 0, 0, d)^T$, $d \neq 0$, does the trick.
