The kernel and image space of $D$ and defining $D$ as a matrix product equation. 
There is an isomorphism between
$P_n(x) = \{p(x) : p(x) = a_0 + a_1x + a_2x^2 +\ldots+ a_nx^n,\ \forall a_i \in \Bbb R\}$
and $\Bbb R^{n+1}$, in the sense that $a_0 + a_1x + a_2x^2 + ... + a_nx^n \in P_n(x)$
may be viewed as $(a_0, a_1, a_2, \ldots , a_n) \in \Bbb R^{n+1}$.
A linear transformation $D : P_n(x) \to P_{n-1}(x)$ such that
$$D[p(x)] = \dfrac{d}{dx}(p(x))$$
a. What is the kernel of $D$;
b. What is the image space of $D$;
c. Define $D$ as a matrix product equation (In the form $A \mathbf v = \mathbf w$). Be sure to specify the domain and the codomain.

I do not particularly understand these concepts, but from what I've read, I've gathered:
The image consists of all the values the function takes in its codomain.
image $D =$ {$f(x) : x \in P_{n-1}(x)$}
The kernel is the set of all zeros of the transformation ie solutions of the equations $A \mathbf v = \mathbf w$ where $\mathbf w = \mathbf 0$
Not sure where to go from here?
Attempted some more.
ker($D$) $=$ {$p(x) \in P_n(x) | D(p(x)) = 0$}
so $p(x)$ is such that
$\dfrac{d}{dx}(p(x)) = 0$
$\dfrac{d}{dx}[a_0 + a_1x + a_2x^2 +...+ a_nx^n] = 0$
$0 + a_1 + 2a_2x +...+ na_nx^{n-1} = 0$
so ker($D$) $=$ {$a_0 + a_1x + a_2x^2 +...+ a_nx^n | 0 + a_1 + 2a_2x +...+ na_nx^{n-1} =  0$}
 A: I'll illustrate these concepts for $n=2$ and let you generalize. Going back to the concept of finding the matrix for a linear transformation, let's first write the matrix of your operator $D$ (with respect to the standard basis on $P_n$), which we will denote by $[D]$:
$$
D(a+bt+ct^2)={d\over dt}(a+bt+ct^2)=b+2ct,
$$
so
$$
\begin{bmatrix} a\\b\\c\end{bmatrix}\overset{D}{\longmapsto}\begin{bmatrix} b\\2c\\0\end{bmatrix}, \tag{$*$}
$$
and therefore 
$$
[D]=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 2\\ 0 & 0 & 0\end{bmatrix}
$$
since
$$
D\left(\begin{bmatrix} a\\b\\c\end{bmatrix}\right)=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 2\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} a\\b\\c\end{bmatrix}=\begin{bmatrix} b\\2c\\0\end{bmatrix}.
$$
So, now to your questions.
(1) Since $\text{ker}(D)=\{p\in P_2:D(p)=0\}$, we only need to consider $[D]\mathbf{v}=\mathbf{0}$, which says $b=0$, $c=0$, and $a$ is free. Hence,
$$
\text{ker}(D)=\left\{\begin{bmatrix} a\\ 0\\ 0\end{bmatrix}:a\in\mathbb{R}\right\}=\{a+0\cdot t+0\cdot t^2:a\in\mathbb{R}\}.
$$
(2) The image space (range space) of $D$ is the set of all possible "outputs" of $D$, and revealed by $(*)$ above:
$$
\text{range}(D)=\left\{\begin{bmatrix} b\\ 2c\\ 0\end{bmatrix}:b,c\in\mathbb{R}\right\}=\{b+2c\cdot t+0\cdot t^2:b,c\in\mathbb{R}\}.
$$
(3) This was answered above by finding $[D]$. Note that $[D]$ is $3\times 3$ since $D$ was a mapping from $P_2$ to $P_2$ which is a three dimensional space.
A: a. Which polynomials in $P_n$ have derivative equal to zero?  (Take the derivative of a general polynomial and see what you need the coefficients to be)
b. What polynomials in $P_{n-1}$ can you get after taking a derivative of a polynomial in $P_n$? (Take the derivative of a general polynomial and see what coefficients you could end up with)
c. Compute $d/dx[a_0+a_1x+...+a_nx^n]$.  This is an element of $P_{n-1}$.  Which element of $R^n$ does it correspond to?  Find the matrix that takes the element of $R^{n+1}$ you started with before taking d/dx, and returns the element of $R^n$ that you found above.
