Consider the $5\times 7$ matrix $A$, \begin{equation*} \begin{bmatrix} 2 & 4 & 1 & -1 & 1 & 4 & 4 \\ 1 & 2 & 1 & 0 & 2 & 4 & 7 \\ 0 & 0 & 1 & 4 & 1 & 8 & 7 \\ 1 & 2 & -1 & 2 & 1 & 9 & 6 \\ -2 & -4 & 1 & 3 & -1 & -2 & -2 \end{bmatrix} \end{equation*}
According to the definition (Definition CSM), the column space of $A$ is \begin{equation*} \csp{A}= \spn{\set{ \colvector{ 2 \\ 1 \\ 0 \\ 1 \\ -2 },\, \colvector{ 4 \\ 2 \\ 0 \\ 2 \\ -4 },\, \colvector{ 1 \\ 1 \\ 1 \\ -1 \\ 1 },\, \colvector{ -1 \\ 0 \\ 4 \\ 2 \\ 3 },\, \colvector{ 1 \\ 2 \\ 1 \\ 1 \\ -1 },\, \colvector{ 4 \\ 4 \\ 8 \\ 9 \\ -2 },\, \colvector{ 4 \\ 7 \\ 7 \\ 6 \\ -2 } }} \end{equation*}
While this is a concise description of an infinite set, we might be able to describe the span with fewer than seven vectors. This is the substance of Theorem BS. So we take these seven vectors and make them the columns of a matrix, which is simply the original matrix $A$ again. Now we row-reduce, \begin{equation*} \begin{bmatrix} 2 & 4 & 1 & -1 & 1 & 4 & 4 \\ 1 & 2 & 1 & 0 & 2 & 4 & 7 \\ 0 & 0 & 1 & 4 & 1 & 8 & 7 \\ 1 & 2 & -1 & 2 & 1 & 9 & 6 \\ -2 & -4 & 1 & 3 & -1 & -2 & -2 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 2 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & \leading{1} & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & \leading{1} & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & \leading{1} & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
The pivot columns are $D=\set{1,\,3,\,4,\,5}$, so we can create the set \begin{equation*} T=\set{ \colvector{ 2 \\ 1 \\ 0 \\ 1 \\ -2 },\, \colvector{ 1 \\ 1 \\ 1 \\ -1 \\ 1 },\, \colvector{ -1 \\ 0 \\ 4 \\ 2 \\ 3 },\, \colvector{ 1 \\ 2 \\ 1 \\ 1 \\ -1 } } \end{equation*} and know that $\csp{A}=\spn{T}$ and $T$ is a linearly independent set of columns from the set of columns of $A$.