Define the linear transformation T: P2 -> R2 by T(p) = [p(0) p(0)] Find a basis for the kernel of T. Pretty lost on how to answer this question. 
Define the linear transformation $T:P_2 \rightarrow \Bbb{R}^2$ by 
$$
T(p) =\left[\begin{array}{c}p(0)\\p(0)\end{array}\right]
$$
Find a basis for the kernel of $T$.
So a $P_2$ polynomial has the form $ax + bx + cx^2$. So $T(p)$ will always have he form $[a\; a]^\intercal$. That would mean the kernel of $T$ is $[a\; a]^\intercal$ where $a = 0$, correct? It was my understanding the kernel of a transformation is all $u$ such that $T(u)= 0$. 
If this is correct,which I'm sure it probably isn't, how do I find the basis? A basis has to be linearly independent and would have to span the kernel, so would it be a polynomial of the form $a + bx + cx^2$? 

If $p \in P_2$, then $p$ has the form $p(x) = ax^2+bx +c$, and $T(x \mapsto ax^2+bx +c) = (c,c)^T$. Hence $T(x \mapsto ax^2+bx +c) = 0 $ iff $c=0$.
It follows that $\ker T = \{ax^2+bx\}_{a,b \in \mathbb{R}} = \operatorname{sp} \{ x \mapsto x, x \mapsto x^2 \}$. Since $x \mapsto x, x \mapsto x^2$ are linearly independent it follows that they are a basis for the kernel.
To see that they are linearly independent, suppose $ax^2+bx = 0$ for all $x$. Differentiating and setting $x=0$ gives $b=0$, then setting $x=1$ gives $a=0$, hence they are linearly independent.
