Archetype V is defined by \begin{equation*} \ltdefn{T}{P_3}{M_{22}},\quad\lt{T}{a+bx+cx^2+dx^3}= \begin{bmatrix} a+b & a-2c\\ d & b-d \end{bmatrix} \end{equation*}
To establish that the linear transformation is surjective, begin by choosing an arbitrary output. In this example, we need to choose an arbitrary $2\times 2$ matrix, say \begin{equation*} \vect{v}=\begin{bmatrix}x&y\\z&w\end{bmatrix} \end{equation*} and we would like to find an input polynomial \begin{equation*} \vect{u}=a+bx+cx^2+dx^3 \end{equation*} so that $\lt{T}{\vect{u}}=\vect{v}$. So we have, \begin{align*} \begin{bmatrix}x&y\\z&w\end{bmatrix}&=\vect{v}\\ &=\lt{T}{\vect{u}}\\ &=\lt{T}{a+bx+cx^2+dx^3}\\ &=\begin{bmatrix} a+b & a-2c\\ d & b-d \end{bmatrix} \end{align*}
Matrix equality leads us to the system of four equations in the four unknowns, $x,y,z,w$, \begin{align*} a+b&=x\\ a-2c&=y\\ d&=z\\ b-d&=w \end{align*} which can be rewritten as a matrix equation, \begin{equation*} \begin{bmatrix} 1 & 1 & 0 & 0\\ 1 & 0 & -2 & 0 \\ 0 & 0 & 0 & 1\\ 0 & 1 & 0 & -1 \end{bmatrix} \colvector{a\\b\\c\\d} = \colvector{x\\y\\z\\w} \end{equation*}
The coefficient matrix is nonsingular, hence it has an inverse, \begin{equation*} \inverse{ \begin{bmatrix} 1 & 1 & 0 & 0\\ 1 & 0 & -2 & 0 \\ 0 & 0 & 0 & 1\\ 0 & 1 & 0 & -1 \end{bmatrix} } = \begin{bmatrix} 1 & 0 & -1 & -1\\ 0 & 0 & 1 & 1\\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\ 0 & 0 & 1 & 0 \end{bmatrix} \end{equation*} so we have \begin{align*} \colvector{a\\b\\c\\d}&= \begin{bmatrix} 1 & 0 & -1 & -1\\ 0 & 0 & 1 & 1\\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2}\\ 0 & 0 & 1 & 0 \end{bmatrix} \colvector{x\\y\\z\\w}\\% &= \colvector{ x-z-w\\ z+w\\ \frac{1}{2}(x-y-z-w)\\ z } \end{align*}
So the input polynomial $\vect{u}=(x-z-w)+(z+w)x+\frac{1}{2}(x-y-z-w)x^2+zx^3$ will yield the output matrix $\vect{v}$, no matter what form $\vect{v}$ takes. This means by Definition SLT that $T$ is surjective. All the same, let us do a concrete demonstration and evaluate $T$ with $\vect{u}$, \begin{align*} \lt{T}{\vect{u}}&=\lt{T}{(x-z-w)+(z+w)x+\frac{1}{2}(x-y-z-w)x^2+zx^3}\\ &= \begin{bmatrix} (x-z-w)+(z+w) & (x-z-w) - 2(\frac{1}{2}(x-y-z-w))\\ z & (z+w)-z \end{bmatrix}\\ &= \begin{bmatrix} x & y\\ z & w \end{bmatrix}\\ &=\vect{v} \end{align*}