Linear transformation via taking derivatives, what are the matrices?

Question

Consider the real-vector space of polynomials (i.e. real coefficients) $f(x)$ of at most degree $3$, let's call that space $X$. And consider the real-vector space of polynomials (i.e. real coefficients) of at most degree $2$, call that $Y$. And consider the linear transformation $A$ from $X$ to $Y$ defined by the following:$$A(f) = 2f' - (x+1)f''.$$Question 1. What's the matrix of $A$ with regards to the bases $\mathfrak{X} = (1,x,x^2,x^3)$ of $X$ and $\mathfrak{Y} = (1, x, x^2)$ of $Y$?

Question 2. What's the matrix of $A$ with regards to the bases $\mathfrak{X} = (1,x+1,(x+1)^2,(x+1)^3)$ of $X$ and $\mathfrak{Y} = (1, x+1, (x+1)^2)$ of $Y$?

I am looking for these two examples I thought of done in detail, given that the examples in my lecture notes are solely with concrete linear algebra things in $\mathbb{R}^n$ and are not in another setting. Seeing them done in detail would give me greater insight on how to move beyond the matrix setting. Thank you!

Answer 1

For a linear transformation, the entire mapping is dependent on the basis elements. Compute A(1), A(x), A(x$^2$), A(x$^3$). The image polynomials will form the columns of A.

Do the same with the second set of bases.

Answer 2

HINT

For question $1$ assume $f(x)=a+bx+cx^2+dx^3$ then

$$A(f)=2f' - (x+1)f''=2(b+2cx+3dx^2)-(x+1)(2c+6dx)=$$$$=2b+4cx+6dx^2-2cx-6dx^2-2c-6dx=(2b-2c)+(2c-6d)x$$

therefore

$$A=\begin{bmatrix}0&2&-2&0\\0&0&2&-6\\0&0&0&0\end{bmatrix}$$

For question $2$ we can proceed in the same way or to perform a change of basis.