Archetype N is the linear transformation \begin{equation*} \ltdefn{T}{\complex{5}}{\complex{3}},\quad \lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}= \colvector{2 x_1 + x_2 + 3 x_3 - 4 x_4 + 5 x_5\\ x_1 - 2 x_2 + 3 x_3 - 9 x_4 + 3 x_5\\ 3 x_1 + 4 x_3 - 6 x_4 + 5 x_5} \end{equation*}
To determine the elements of $\complex{3}$ in $\rng{T}$, find those vectors $\vect{v}$ such that $\lt{T}{\vect{u}}=\vect{v}$ for some $\vect{u}\in\complex{5}$, \begin{align*} \vect{v}&=\lt{T}{\vect{u}}\\ &= \colvector{ 2 u_1 + u_2 + 3 u_3 - 4 u_4 + 5 u_5\\ u_1 - 2 u_2 + 3 u_3 - 9 u_4 + 3 u_5\\ 3 u_1 + 4 u_3 - 6 u_4 + 5 u_5}\\ &= \colvector{2u_1\\u_1\\3u_1}+ \colvector{u_2\\-2u_2\\0}+ \colvector{3u_3\\3u_3\\4u_3}+ \colvector{-4u_4\\-9u_4\\-6u_4}+ \colvector{5u_5\\3u_5\\5u_5}\\ &= u_1\colvector{2\\1\\3}+ u_2\colvector{1\\-2\\0}+ u_3\colvector{3\\3\\4}+ u_4\colvector{-4\\-9\\-6}+ u_5\colvector{5\\3\\5}\\ \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 five vectors \begin{align*} \colvector{2\\1\\3}&& \colvector{1\\-2\\0}&& \colvector{3\\3\\4}&& \colvector{-4\\-9\\-6}&& \colvector{5\\3\\5} \end{align*} using the scalars $u_1,\,u_2,\,u_3,\,u_4,\,u_5$. Furthermore, since $\vect{u}$ can be any element of $\complex{5}$, every such linear combination is an output. This means that \begin{equation*} \rng{T}=\spn{\set{ \colvector{2\\1\\3},\, \colvector{1\\-2\\0},\, \colvector{3\\3\\4},\, \colvector{-4\\-9\\-6},\, \colvector{5\\3\\5} }} \end{equation*}
The five vectors in this spanning set for $\rng{T}$ form a linearly dependent set (Theorem MVSLD). 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 (nice) basis, \begin{equation*} \rng{T}=\spn{\set{ \colvector{1\\0\\0},\,\colvector{0\\1\\0},\,\colvector{0\\0\\1} } }=\complex{3}\end{equation*}