Suppose in the course of analyzing a matrix (its column space, its null space, its …) we encounter the following set of vectors, described by a span \begin{equation*} X=\spn{\set{ \colvector{1\\2\\1\\6\\6},\, \colvector{3\\-1\\2\\-1\\6},\, \colvector{1\\-1\\0\\-1\\-2},\, \colvector{-3\\2\\-3\\6\\-10} }} \end{equation*}
Let $A$ be the matrix whose rows are the vectors in $X$, so by design $X=\rsp{A}$, \begin{equation*} A= \begin{bmatrix} 1 & 2 & 1 & 6 & 6\\ 3 & -1 & 2 & -1 & 6\\ 1 & -1 & 0 & -1 & -2\\ -3 & 2 & -3 & 6 & -10 \end{bmatrix} \end{equation*}
Row-reduce $A$ to form a row-equivalent matrix in reduced row-echelon form, \begin{equation*} B= \begin{bmatrix} \leading{1} & 0 & 0 & 2 & -1\\ 0 & \leading{1} & 0 & 3 & 1\\ 0 & 0 & \leading{1} & -2 & 5\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}
Then Theorem BRS says we can grab the nonzero columns of $\transpose{B}$ and write \begin{equation*} X=\rsp{A}=\rsp{B}= \spn{\set{ \colvector{1\\0\\0\\2\\-1},\, \colvector{0\\1\\0\\3\\1},\, \colvector{0\\0\\1\\-2\\5} }} \end{equation*}
These three vectors provide a much-improved description of $X$. There are fewer vectors, and the pattern of zeros and ones in the first three entries makes it easier to determine membership in $X$.