Basis for Kernel in function space Here is my problem:
Let $\Pi_2$ be the function space of all polynomials of max. grade 2,that are defined on the interval $[-1,1]$.
Let $G : \Pi_2 \to \mathbb{R}$ be given as $f \stackrel{G}{\mapsto} f(1) $. 
What is the basis of the $\ker(G)$ ?

My thoughts process is to find the set of the kernel and then find the basis to it.
The kernel is the set of functions which evaluate to zero. So:
$\ker(G):= \{ \ f \  | \ f(1) = 0\}  $ I dont know how to write this set more formally.
My next step would be to find the basis to this set.
https://en.wikipedia.org/wiki/Kernel_(linear_algebra)#Examples
 A: First$\DeclareMathOperator{\Ima}{Im}$$\DeclareMathOperator{\Ker}{Ker}$ notice that $\Ima G = \mathbb{R}$ since $G(c) = c$, for every $c \in \mathbb{R}$. Now we can use the rank-nullity theorem to determine the dimension of $\Ker G$.
$$3 = \dim \Pi_2 = \dim \Ima G + \dim \Ker G = 1 + \dim\Ker G \implies \dim\Ker G = 2$$
Now find two simple linearly independent polynomials in $\Ker G$. For example, you can take $x-1$ and $(x-1)^2$. They are obviously not proportional since their degrees are not equal.
Thus, the set $\{x-1, (x-1)^2\} \subseteq \Ker G$ is a linearly independent set with cardinality equal to the dimension of the space. Hence, it is a basis for $ \Ker G$.
A: Here is another take to complement @mechanodroid's excellent answer.
If $p \in \ker G$ we must have $p(1) = 0$ and so $x-1$ must be a factor
of the polynomial. Similarly, if $x-1$ is a factor  of  a polynomial $p$
then we must have $p \in \ker G$.
Hence $\ker G = \{ p \in \Pi_2 | x-1 \text{ is a factor} \}
= \{ x \mapsto (x-1) q(x) | g \in \Pi_1 \}$ .
We can quickly check that the linear map $L :\Pi_1 \to \ker G$ given by
$(Lq)(x) = (x-1)q(x)$ is a bijection, hence $\dim \ker G = \dim \Pi_1 = 2$.
Furthermore, since $x \mapsto 1, x \mapsto  x$ forms a basis for
$\Pi_1$, then $L (x \mapsto 1), L(x \mapsto  x)$ forms a basis for
$\ker G$.
Hence $x \mapsto x-1, x \mapsto x(x-1)$ is a suitable basis.
