Kernel and Image of a polynomial linear transformation I have here a linear transformation $T : P_3(\mathbb{R})\rightarrow P_3(\mathbb{R}) $ defined by:

$ T(at^3 + bt^2 + ct + d) = (a-b)t^3 + (c-d)t $

I'm very very new in this subject and I'm not going well with polynomials. I need find the $ Kernel $ and the $ Image $ of the transformation. Look what I've been thinking:

$Ker(T) =  \{ T(p) = 0 / p \in P_3\} $
$ T(at^3 + bt^2 + ct + d) = (a-b)t^3 + (c-d)t = 0 $
$(a-b) = 0 \ ;\ \ (c-d) = 0 \ ;\ \ a = b \ ; \ \ c = d $
$ Ker(T) = \{ at^3 + at^2 + ct +c\ /\ a,c \in \mathbb{R} \} $

And what about the $ Image $? I know that $Im(T) = \{ T(p) / p \in P_3 \}$, but how can I show it? And how can I test if a polynomial such as $ p(t) = t^3 + t^2 + t -1 \in Im(t)$?
 A: 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*}
A: We can set up the matrix of the linear transformation  $T:P_3(\mathbb{R})\rightarrow P_3(\mathbb{R})$, then find its null space and column space, respectively. First, if we agree to represent the third-order polynomial $P_3=at^3 + bt^2 + ct + d$ by the column vector $\begin{pmatrix}a &b& c& d\end{pmatrix}^T$, then
$$T=\begin{pmatrix}
1 & -1 & 0 &0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & -1\\
0 & 0 & 0 & 0
\end{pmatrix}.$$
It is obvious that the pivots of $T$ is on the first and the third columns, so the kernel of T (i.e., the null space) is 
$$\mathrm{ker}\, T=\mathrm{span}\{\begin{pmatrix} 1\\1\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\0\\1\\1 \end{pmatrix}\},$$
and the image of $T$ is spanned by the pivot columns of $T$:
$$\mathrm{im}\, T=\mathrm{span}\{\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\0\\1\\0 \end{pmatrix}\}.$$
The answers are identical to what have been given. This is just to show we can get the answers by a slightly different approach.
