The kernel is correct. Additionally, since the kernel depends on only two coefficients $a$ and $c$, it has dimension 2.

**For the image:**

Take any polynomial $p(t)=At^3+Bt^2+Ct+E$.

The question now is: How do $A,B,C,E$ have to look for there to exist some $a,b,c,d$ such that $T(at^3+bt^2+ct+d)=p(t)$?

The question is equivalent to solving for $A, B, C, E$ in the equation: $(a-b)t^3+0t^2+(c-d)t+0=At^3+Bt^2+Ct+E$.

We now have:

$A=a-b$,

$B=0$,

$C=c-d$

$E=0$

We can take:

$a=A$,

$b=0$,

$c=C$

$d=0$

Consequently, $p$ is in the image, iff $B=0=E$. The image, then, is:
\begin{align*}
\mbox{Im}(T)=\{At^3+Ct\ |\ A,C\in\mathbb R\}.
\end{align*}

As soon as you have the matrix of a linear map $f$, you have a system of generators of the image of $f$.

Indeed the column-vectors of the matrix are the images of the vectors of the basis in the source-space, say $v_1,v_2,v_3$. Hence for any vector $v$ in $\mathbf R^3$, $v=\lambda_1 v_1+\lambda_2v_2+\lambda_3v_3$, we have:
$$f(v)=\lambda_1 f(v_1)+\lambda_2f(v_2)+\lambda_3f(v_3).$$

Naturally, this system of generators is not *minimal*, i. e. it is not a basis of the image in general. But from this system, you can deduce a basis by column reduction.

It will not be necessary here, because the rank-nullity theorem ensures $\dim\operatorname{Im}f=2$ since you've found that $\dim\ker f=1$. Thus, $f$ is surjective, i. e. the image of $f$ is the whole of $\mathbf R^2$.

## Best Answer

Let us see what happens to $p_3(x) = a_3x^3 + a_2 x^2 + a_1x + a_0$ after derivation. You get $p'(x)=3a_3 x^2 + 2a_2 x + a_1$. Namely, the original polynomial $p_3(x)$ is spanned by $\{x^3, x^2, x, 1\}$ where the Image is spanned by $\{x^2, x, 1\}$. As such, the polynomials which derivation is $0$ are constants, thus $KerT = sp\{1 \}$. You can easily check that $\dim P_3 = \dim P_2 + \dim P_0. $