= Solution
The pivot columns in the <row echelon form> are columns 1 and 2. Take the corresponding columns of the \b[original] <matrix>:
$$
\boxed{\operatorname{Col}(A)\text{ has basis }\left\{\begin{pmatrix}1\\2\\2\\0\end{pmatrix},\begin{pmatrix}1\\3\\0\\1\end{pmatrix}\right\}.}
$$
For a direct check, call these columns $c_1,c_2$. The other columns satisfy $c_3=c_1-c_2$ and $c_4=c_1+c_2$, so they are in the same <linear span>. Also $c_1,c_2$ are <linearly independent>: their fourth entries force the coefficient of $c_2$ to vanish, then their first entries force the other coefficient to vanish. <Elementary row operations> preserve dependence relations among columns, but generally change the <column space> itself, which is why the original columns must be used.
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