kernel and nullity of $T: P_2 \rightarrow \mathbb{R}^2$ with $T(p(x)) = \begin{bmatrix}p(0)\\p(1)\end{bmatrix}$ I am not 100% sure on this question:

Find the nullity of $T: P_2 \rightarrow \mathbb{R}^2$ with $T(p(x)) =
 \begin{bmatrix}p(0)\\p(1)\end{bmatrix}$

First I tried to find $\ker T$, which I think is:
$\ker T = \{0\}$
as $p(0) = 0 \Rightarrow a = 0$ in $a + bx + cx^2$
Following this, I concluded the dimension of the kernel = nullity T = 0.
Is my thought process correct?
EDIT: as moonlight stated, the nullity is 1.
I failed to recognize following $a=0$ in $a + bx + cx^2$ that:
$p(1) = 0 \Rightarrow b = -c$
so $\ker T = \{bx -bx^2 | b\in \mathbb{R}\}$
 A: No. If $p(x)\in\ker T$, then $p(0)=p(1)=0$. So $p(x)=ax(x-1)$ by Factor Theorem.
So $\dim\ker T=\dim\{ax(x-1)\mid a\in\Bbb R\}=1$, so nullity of $T$ is 1.
A: $T(p(x)) =
 \begin{bmatrix}p(0)\\p(1)\end{bmatrix}$
Now basis elements of $P_2$ are
$P_2 = \{1\ ,x\ ,x²\}$
Now ,
$T(1) =\begin{bmatrix}1\\1\end{bmatrix}$= 1$\begin{bmatrix}1\\0\end{bmatrix}$+ 1$\begin{bmatrix}0\\1\end{bmatrix}$
$T(x) =\begin{bmatrix}0\\1\end{bmatrix}$= 0$\begin{bmatrix}1\\0\end{bmatrix}$+ 1$\begin{bmatrix}0\\1\end{bmatrix}$
$T(x²) =\begin{bmatrix}0\\1\end{bmatrix}$= 0$\begin{bmatrix}1\\0\end{bmatrix}$+ 1$\begin{bmatrix}0\\1\end{bmatrix}$
Now the matrix representation of the transformation is given by $\begin{bmatrix}1 & 0 &0\\ 1 & 1 & 1\end{bmatrix}$
Now finding the basis elements of N(T)= solution space of AX=0
$\begin{bmatrix}1 & 0 & 0\\ 1 & 1 & 1\end{bmatrix}$ $\begin{bmatrix}x\\y\end{bmatrix}$=$\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow\left[\begin{array}{ccc|c}1 & 0 & 0 & 0\\1 & 1 & 1 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc|c}1 & 1 & 1 & 0\\0 & -1 & -1 & 0\end{array}\right]$
Now here rank of the matrix is $2$
And by rank nullity theorem ,
$\Rightarrow\ 3= Rank(T)+Nullity (T)$
$\Rightarrow\ 3= 2 +Nullity (T)$
$\Rightarrow\ Nullity (T) = 1$
