Begin with a set of five vectors from $\complex{4}$, \begin{equation*} S=\set{ \colvector{ 1 \\ 1 \\ 2 \\ 1},\, \colvector{ 2 \\ 2 \\ 4 \\ 2},\, \colvector{ 2 \\ 0 \\ -1 \\ 1},\, \colvector{ 7 \\ 1 \\ -1 \\ 4},\, \colvector{ 0 \\ 2 \\ 5 \\ 1} } \end{equation*} and let $W=\spn{S}$. To arrive at a (smaller) linearly independent set, follow the procedure described in Theorem BS. Place the vectors from $S$ into a matrix as columns, and row-reduce, \begin{equation*} \begin{bmatrix} 1 & 2 & 2 & 7 & 0 \\ 1 & 2 & 0 & 1 & 2 \\ 2 & 4 & -1 & -1 & 5 \\ 1 & 2 & 1 & 4 & 1 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 2 & 0 & 1 & 2 \\ 0 & 0 & \leading{1} & 3 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
Columns 1 and 3 are the pivot columns ($D=\set{1,\,3}$) so the set \begin{equation*} T=\set{ \colvector{ 1 \\ 1 \\ 2 \\ 1},\, \colvector{ 2 \\ 0 \\ -1 \\ 1} } \end{equation*} is linearly independent and $\spn{T}=\spn{S}=W$. Boom!
Since the reduced row-echelon form of a matrix is unique (Theorem RREFU), the procedure of Theorem BS leads us to a unique set $T$. However, there is a wide variety of possibilities for sets $T$ that are linearly independent and which can be employed in a span to create $W$. Without proof, we list two other possibilities: \begin{align*} T^{\prime}&=\set{ \colvector{ 2 \\ 2 \\ 4 \\ 2},\, \colvector{ 2 \\ 0 \\ -1 \\ 1} }\\ T^{*}&=\set{ \colvector{3 \\ 1 \\ 1 \\ 2},\, \colvector{-1 \\ 1 \\ 3 \\ 0} } \end{align*}
Can you prove that $T^{\prime}$ and $T^{*}$ are linearly independent sets and $W=\spn{S}=\spn{T^{\prime}}=\spn{T^{*}}$?