Define the image $T:P_2\rightarrow \mathbb{R}^2$ by $T(p)=\begin{pmatrix}p(0) \\ p(1)\end{pmatrix}.$
a) Find one polynomial $p\in P_2$ that spans the nullspace $N(T).$
b) Describe the value space $V(T).$
a) Since we have a polynomial in of 2nd degree in $\mathbb{R}^2$, I denoted
$$p(x)=a_0+a_1x+a_2x^2 \implies \left\{ \begin{array}{rcr} p(0) & = & a_0 \\ p(1) & = & a_0+a_1+a_2 \\ \end{array} \right.$$
In order to get the nullspace, I need to solve the homogenous equation
$$\begin{pmatrix}p(0) \\ p(1)\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix} \Leftrightarrow \left\{ \begin{array}{rcr} a_0 & = & 0\\ a_0+a_1+a_2 & = & 0 \\ \end{array} \right. \sim a_1+a_2=0. $$
Setting $a_1=t$ I get $a_2=-t$ and $a_0=0$. So the nullspace is the line $t(0,1,-1)^T.$
How do I find the polynomial that spans this nullspace?
b) The answer to this is $V(T)=\mathbb{R}^2.$ Can someone explain to me why this is?
A polynomial $p$ is in the null space if $p(0) =p(1) = 0$. This means that $x(x-1)$ divides $p$ and if it divides $p$ it's in the null space. So $q(x) = x(x-1)$ spans the null space. As $q(x)= -x + x^2$ this agrees (sort of) with your answer.
$3= \dim(P_2) = \dim(N(T)) + \dim(R(T)) = 1 + \dim(R(T)$ by the standard dimension theorem. It follows that $R(T)$ is two-dimensional and so $T$ must be surjective.