When we first defined the span of a set of column vectors, in Example SCAD we looked at the set \begin{equation*} W=\spn{\set{ \colvector{2\\-3\\1},\, \colvector{1\\4\\1},\, \colvector{7\\-5\\4},\, \colvector{-7\\-6\\-5} }} \end{equation*} with an eye towards realizing $W$ as the span of a smaller set. By building relations of linear dependence (though we did not know them by that name then) we were able to remove two vectors and write $W$ as the span of the other two vectors. These two remaining vectors formed a linearly independent set, even though we did not know that at the time.
Now we know that $W$ is a subspace and must have a basis. Consider the matrix, $C$, whose rows are the vectors in the spanning set for $W$, \begin{equation*} C=\begin{bmatrix} 2 & -3 & 1\\ 1 & 4 & 1\\ 7 & -5 & 4\\ -7 & -6 & -5 \end{bmatrix} \end{equation*}
Then, by Definition RSM, the row space of $C$ will be $W$, $\rsp{C}=W$. Theorem BRS tells us that if we row-reduce $C$, the nonzero rows of the row-equivalent matrix in reduced row-echelon form will be a basis for $\rsp{C}$, and hence a basis for $W$. Let us do it — $C$ row-reduces to \begin{equation*} \begin{bmatrix} \leading{1} & 0 & \frac{7}{11}\\ 0 & \leading{1} & \frac{1}{11}\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \end{equation*}
If we convert the two nonzero rows to column vectors then we have a basis, \begin{equation*} B=\set{\colvector{1\\0\\\frac{7}{11}},\,\colvector{0\\1\\\frac{1}{11}}} \end{equation*} and \begin{equation*} W=\spn{\set{\colvector{1\\0\\\frac{7}{11}},\,\colvector{0\\1\\\frac{1}{11}}}} \end{equation*}
For aesthetic reasons, we might wish to multiply each vector in $B$ by $11$, which will not change the spanning or linear independence properties of $B$ as a basis. Then we can also write \begin{equation*} W=\spn{\set{\colvector{11\\0\\7},\,\colvector{0\\11\\1}}} \end{equation*}