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 $\complex{3}$ in $\krn{T}$, find those vectors $\vect{u}$ such that $\lt{T}{\vect{u}}=\zerovector$, that is, \begin{align*} \lt{T}{\vect{u}}&=\zerovector\\ \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{0\\0\\0\\0\\0} \end{align*}
Vector equality (Definition CVE) leads us to a homogeneous system of 5 equations in the variables $u_i$, \begin{align*} -u_1 + u_2 - 3 u_3&=0\\ -u_1 + 2 u_2 - 4 u_3&=0\\ u_1 + u_2 + u_3&=0\\ 2 u_1 + 3 u_2 + u_3&=0\\ u_1 + 2 u_3&=0 \end{align*}
Row-reducing the coefficient matrix gives \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 2\\ 0 & \leading{1} & -1\\ 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} \end{equation*}
The kernel of $T$ is the set of solutions to this homogeneous system of equations, which by Theorem BNS can be expressed as \begin{equation*} \krn{T}=\spn{\set{\colvector{-2\\1\\1}} }\end{equation*}