Archetype O is the linear transformation \begin{equation*} \ltdefn{T}{\complex{3}}{\complex{5}},\quad \lt{T}{\colvector{x_1\\x_2\\x_3}}= \colvector{-x_1 + x_2 - 3 x_3\\ -x_1 + 2 x_2 - 4 x_3\\ x_1 + x_2 + x_3\\ 2 x_1 + 3 x_2 + x_3\\ x_1 + 2 x_3 } \end{equation*}
To determine the elements of $\complex{5}$ in $\rng{T}$, find those vectors $\vect{v}$ such that $\lt{T}{\vect{u}}=\vect{v}$ for some $\vect{u}\in\complex{3}$, \begin{align*} \vect{v}&=\lt{T}{\vect{u}}\\ &=\colvector{-u_1 + u_2 - 3 u_3\\ -u_1 + 2 u_2 - 4 u_3\\ u_1 + u_2 + u_3\\ 2 u_1 + 3 u_2 + u_3\\ u_1 + 2 u_3 }\\ &= \colvector{-u_1\\-u_1\\u_1\\2 u_1\\ u_1} + \colvector{u_2\\2u_2\\u_2\\3u_2\\ 0} + \colvector{-3u_3\\-4u_3\\u_3\\u_3\\ 2 u_3}\\ &= u_1\colvector{-1\\-1\\1\\2\\1} + u_2\colvector{1\\2\\1\\3\\ 0} + u_3\colvector{-3\\-4\\1\\1\\2} \end{align*}
This says that every output of $T$ (in other words, the vector $\vect{v}$) can be written as a linear combination of the three vectors \begin{align*} \colvector{-1\\-1\\1\\2\\1} && \colvector{1\\2\\1\\3\\ 0} && \colvector{-3\\-4\\1\\1\\2} \end{align*} using the scalars $u_1,\,u_2,\,u_3$. Furthermore, since $\vect{u}$ can be any element of $\complex{3}$, every such linear combination is an output. This means that \begin{equation*} \rng{T}=\spn{\set{ \colvector{-1\\-1\\1\\2\\1},\, \colvector{1\\2\\1\\3\\ 0},\, \colvector{-3\\-4\\1\\1\\2} }} \end{equation*}
The three vectors in this spanning set for $\rng{T}$ form a linearly dependent set (check this!). So we can find a more economical presentation by any of the various methods from Section CRS and Section FS. We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to Theorem BRS, so we can describe the range of $T$ with a basis, \begin{equation*} \rng{T}=\spn{\set{ \colvector{1\\0\\-3\\-7\\-2},\,\colvector{0\\1\\2\\5\\1} } }\end{equation*}