Archetype V is the linear transformation \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*}
Define the function $\ltdefn{S}{M_{22}}{P_3}$ defined by \begin{equation*} \lt{S}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}=(a - c - d)+ (c + d)x +\frac{1}{2}(a - b - c - d)x^2+cx^3 \end{equation*}
Then \begin{align*} &\lt{\left(\compose{T}{S}\right)}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}\\ &\quad\quad=\lt{T}{\lt{S}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}}\\ &\quad\quad=\lt{T}{(a - c - d)+ (c + d)x +\frac{1}{2}(a - b - c - d)x^2+cx^3}\\ &\quad\quad=\begin{bmatrix} (a - c - d)+ (c + d)&(a - c - d)-2(\frac{1}{2}(a - b - c - d))\\c&(c + d)-c \end{bmatrix}\\ &\quad\quad=\begin{bmatrix}a&b\\c&d\end{bmatrix}\\ &\quad\quad=\lt{I_{M_{22}}}{\begin{bmatrix}a&b\\c&d\end{bmatrix}} \end{align*} and \begin{align*} &\lt{\left(\compose{S}{T}\right)}{a+bx+cx^2+dx^3}\\ &\quad\quad=\lt{S}{\lt{T}{a+bx+cx^2+dx^3}}\\ &\quad\quad=\lt{S}{\begin{bmatrix} a+b&a-2c\\d&b-d \end{bmatrix}}\\ &\quad\quad=((a+b)-d-(b-d))+ (d+(b-d))x\\ &\quad\quad+\left(\frac{1}{2}((a+b)-(a-2c)-d-(b-d))\right)x^2+ (d)x^3\\ &\quad\quad=a+bx+cx^2+dx^3\\ &\quad\quad=\lt{I_{P_3}}{a+bx+cx^2+dx^3} \end{align*}
For now, understand why these computations show that $T$ is invertible, and that $S=T^{-1}$. Maybe even be amazed by how $S$ works so perfectly in concert with $T$! We will see later just how to arrive at the correct form of $S$ (when it is possible).