Answer
See below
Work Step by Step
Obtain: $Ker (T)=\{p \in P_2(R): T(p)=0\}\\
=\{\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}:(a-b+d)+x^2(-a+b-d)+bx^2=0\}$
We have the system:
$a-b+d=0\\
-a+b-d=0$
thus, $a=b-d$
Thus, $Ker (T)=\{\begin{bmatrix}
b-d & b\\
c & d
\end{bmatrix}:b,c,d\in R\}\\
=\{\begin{bmatrix}
b & b\\
0 & 0
\end{bmatrix}+\begin{bmatrix}
0 & 0\\
c & 0
\end{bmatrix}+\begin{bmatrix}
-d & 0\\
0& d
\end{bmatrix}:b,c,d \in R\}\\
=\{b\begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix}+c\begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix}+d\begin{bmatrix}
-1 & 0\\
0& 1
\end{bmatrix}:b,c,d \in R\}\\
=span\{\begin{bmatrix}
1 & 1\\
0 & 0
\end{bmatrix};\begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix};\begin{bmatrix}
-1 & 0\\
0& 1
\end{bmatrix}\}$
$\dim [Ker(T)]=3$
Apply Rank Nullity Theorem:
$Rng(T)=\{T(A): A \in M_2(R)\}\\
=\{(a-b+d)+(-a+b-d)x^2: a,b,c \in R\}\\
=\{(a-b+d)-(a-b+d)x^2: a,b,d \in R\}\\
=\{(a-b+d)(1-x^2):a,b,d \in R\}\\
=span\{1-x^2\}$
Since $\{1-x^2\}$ is the basic for $Rng(T)$,
then $\dim [Rng(T)]=1$
