What is the form of bases in a non square matrix? I am trying to find the null space and the column space for this matrix.
$\pmatrix{1&1&1&1\\1&2&0&3\\0&1&-1&2}$
The process I usually follow is to first convert to REF, which gives:
$\pmatrix{1&0&2&-1\\0&1&-1&2\\0&0&0&0}$
Which, reading from REF, seems to give a basis of the form $\pmatrix{-2\\1\\1\\0}\pmatrix{1\\-2\\0\\1}$ for the null space. It looks a little odd, but I suppose it makes sense that a matrix with $n=4$ would have vectors of $4$ entries.
Where I am really confused is how to find a basis for col(A). I would usually just take the two pivot columns and call that a basis, but clearly those vectors would not be the same size as the ones in null(A). How else would I find col(A) then?
 A: In general, the null space and column space of a matrix don't have to be the same. By definition, the column space of a matrix is the span of its columns; since your matrix here has 3 rows, all of its columns are vectors in $\mathbb{R}^3$. As such, its basis should not look like your null space basis. The same procedure you describe, taking the pivot columns, suffices to give you a basis.
A: Column space is:
$\left\{ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \right\}$
Those are the columns from the original matrix that are pivot columns in the reduced form.
A: You have the correct null space.
The easiest way to find the column space, it to take the maximal number of independent columns.
If we add the second and third columns we get a scalar multiple of the first column.  Choose any 2 of the first 3 columns.  The 4th column is also not linearly independent.
Any two columns will suffice:
$\begin{bmatrix} 1\\1\\0\end{bmatrix},\begin{bmatrix} 1\\2\\1\end{bmatrix}$
But you could just as easily take 
$\begin{bmatrix} 1\\0\\-1\end{bmatrix},\begin{bmatrix} 1\\3\\2\end{bmatrix}$
A: Let $R$ be the REF (Row Echelon Form) form of matrix $A$.
Since row operations preserve the nullspace, we have
$Ax=0\Leftrightarrow Rx=0$.
Let $x=(x_1\,x_2\,x_3\,x_4), A=(a_1|a_2|a_3)$ and $R=(r_1|r_2|r_3)$.
Both here and in your problem (3 rows indicates $\mathbb{R}^3$ because of 3 coordinates like x,y,z) $a_i,r_i\in \mathbb{R}^3$.
Notice the following:
$Ax=0\leftrightarrow x_1a_1+x_2a_2+x_3a_3+x_4a_4=0$ and
$Rx=0\leftrightarrow x_1r_1+x_2r_2+x_3r_3+x_4r_4=0$.
The two long equations tell us that row operations preserve linear combination relations.
Since the the last two columns of $R$ are linear combinations of the first two columns, the same also holds in $A$. Therefore, to generate the column space of $A$, we just need the first two columns of $A$. So they are (one of the) basis of $col(A)$.
