= Solution
Let $c_1,\ldots,c_4$ be the original columns. The first three are <linearly independent>: restricting them to rows 2, 3 and 4 gives the triangular <matrix>
$$
\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix},
$$
whose <determinant> is one. Also $c_4=c_1-c_2+c_3$, so a <basis> for the <column space> is
$$
\boxed{\mathcal B_{\rm col}=\left\{
\begin{pmatrix}1\\1\\0\\0\\1\end{pmatrix},
\begin{pmatrix}1\\1\\1\\0\\0\end{pmatrix},
\begin{pmatrix}1\\0\\1\\1\\0\end{pmatrix}\right\}.}
$$
These are columns of the original <matrix>. Taking columns of its <reduced row echelon form> would generally give the wrong <column space>.
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