Define a linear transformation $\ltdefn{T}{P_3}{M_{22}}$ by \begin{equation*} \lt{T}{a+bx+cx^2+dx^3}=\begin{bmatrix}a+b&a-2c\\d&b-d\end{bmatrix} \end{equation*}
We verify the two defining conditions of a linear transformation. \begin{align*} \lt{T}{\vect{x}+\vect{y}}&= \lt{T}{(a_1+b_1x+c_1x^2+d_1x^3)+(a_2+b_2x+c_2x^2+d_2x^3)}\\ &=\lt{T}{(a_1+a_2)+(b_1+b_2)x+(c_1+c_2)x^2+(d_1+d_2)x^3}\\ &=\begin{bmatrix} (a_1+a_2)+(b_1+b_2)&(a_1+a_2)-2(c_1+c_2)\\ d_1+d_2&(b_1+b_2)-(d_1+d_2) \end{bmatrix}\\ &=\begin{bmatrix} (a_1+b_1)+(a_2+b_2)&(a_1-2c_1)+(a_2-2c_2)\\ d_1+d_2&(b_1-d_1)+(b_2-d_2) \end{bmatrix}\\ &=\begin{bmatrix}a_1+b_1&a_1-2c_1\\d_1&b_1-d_1\end{bmatrix}+ \begin{bmatrix}a_2+b_2&a_2-2c_2\\d_2&b_2-d_2\end{bmatrix}\\ &=\lt{T}{a_1+b_1x+c_1x^2+d_1x^3}+\lt{T}{a_2+b_2x+c_2x^2+d_2x^3}\\ &=\lt{T}{\vect{x}}+\lt{T}{\vect{y}} \end{align*} and \begin{align*} \lt{T}{\alpha\vect{x}}&=\lt{T}{\alpha(a+bx+cx^2+dx^3)}\\ &=\lt{T}{(\alpha a)+(\alpha b)x+(\alpha c)x^2+(\alpha d)x^3}\\ &=\begin{bmatrix} (\alpha a)+(\alpha b)&(\alpha a)-2(\alpha c)\\ \alpha d&(\alpha b)-(\alpha d) \end{bmatrix}\\ &=\begin{bmatrix} \alpha(a+b)&\alpha(a-2c)\\ \alpha d&\alpha(b-d) \end{bmatrix}\\ &=\alpha\begin{bmatrix}a+b&a-2c\\d&b-d\end{bmatrix}\\ &=\alpha\lt{T}{a+bx+cx^2+dx^3}\\ &=\alpha\lt{T}{\vect{x}} \end{align*}
So by Definition LT, $T$ is a linear transformation.