= Solution
The <pivot columns> in the <row echelon form> are columns 1 and 3. Use those columns of the \b[original] <matrix> for a <basis> of the original <column space>:
$$
\boxed{\left\{\begin{pmatrix}1\\1\\-1\end{pmatrix},\begin{pmatrix}-1\\4\\6\end{pmatrix}\right\}.}
$$
For a direct spanning check, column 2 is twice column 1 and column 4 is $(17/5)$ times column 1 minus $(3/5)$ times column 3. The displayed columns are <linearly independent>, since the first two coordinates of $\alpha c_1+\beta c_3=0$ give $\alpha-\beta=0$ and $\alpha+4\beta=0$, so $\alpha=\beta=0$.
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