Kernel, Range, and Matrix Representation of a Linear Transformation Let L be defined on $P_3$ (the vector space of polynomials of degree less than 3) by
$L(p) = q$ where $q(x) = 4p(x) − 3xp'(x) + x^2 p''(x)$.
(a) Find the range of L in the form Span(. . .).
(b) Find the the kernel of L in the form Span(. . .).
(c) Find all solutions for the equation $L(p) = 16 − x$.
(d) Find the matrix representation of $L : P_3 → P_3$ with respect to the ordered basis $[1, x, x^2]$.
I'm just looking for someone to tell me if what I have for a-c is correct as well as help we with part d!
Here is what I have for the first 3 parts:
After reducing and combining like terms, $L(p) = 7bx + 4c$
a) Range of L = $span\{1,x\}$
b) $ker(L) = span\{x^2\}$
c) $7bx + 4c = 16 - x$
$b = -1/7, c = 4$
d) ????
Again I'm not sure if these answers are correct. So thank you for your help!
 A: It easy/efficient to start with (d), then answer the others. 
First note that upon simplifying the given $L(p(x))=q(x)$, the mapping here is $$L(a+bx+cx^2)=4a+bx,$$ which can be written in terms of coordinates of the given ordered basis on $P_3$ as
$$L\begin{bmatrix} a\\ b\\ c\end{bmatrix}=\begin{bmatrix} 4a\\ b\\ 0\end{bmatrix}.$$ Thus, the matrix of the transformation $L$ (with respect to the given basis) is $M_L$ where
$$
\underbrace{\begin{bmatrix} 4 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}}_{M_L}\ \underbrace{\begin{bmatrix}a\\ b\\ c\end{bmatrix}}_\text{input}=\underbrace{\begin{bmatrix}4a\\ b\\ 0\end{bmatrix}}_\text{output}.\tag{1}$$

Now, the answer to (a) is evident from $(1)$: $$\text{range}(L)=\left\{\begin{bmatrix}4a\\ b\\ 0\end{bmatrix}\right\}=\{4a+bx\}=\text{span}\{1,x\}.$$

Similarly, for (b), $\ker(L)=\{v:Lv=0\}$ so
$$
\begin{bmatrix} 4 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}a\\b\\ c\end{bmatrix}=\begin{bmatrix}0\\0\\ 0\end{bmatrix} \implies \begin{bmatrix} 4 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}$$
so $a=0$, $b=0$, and $c$ is free. Thus,
$$
\ker(L)=\{cx^2\}=\text{span}\{x^2\}.
$$

Finally for (c), we want to solve
$$
\begin{bmatrix} 4 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}a\\b\\ c\end{bmatrix}=\begin{bmatrix}16\\-1\\ 0\end{bmatrix} \implies \begin{bmatrix} 4 & 0 & 0 & 16\\ 0 & 1 & 0 & -1\\ 0 & 0 & 0 & 0\end{bmatrix}\rightarrow \begin{bmatrix} 1 & 0 & 0 & 4\\ 0 & 1 & 0 & -1\\ 0 & 0 & 0 & 0\end{bmatrix}$$
so $a=4$, $b=-1$, and $c$ is free. Thus, the solution we seek is $\{4-x+cx^2\}$.
