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 $a + bx + cx^2$? 
 A: Your guess is that the kernel is $\left[\begin{matrix}a\\a\end{matrix}\right]$, but that can't be right, because it is not an element of $P_2$.
The kernel is all the polynomials $p(x)$ of degree $\leq 2$ such that $p(0)=0$, that is, polynomials of the form $bx + ax^2$, for any $a,b$ in the real numbers. So we need to find a set of polynomials such that multiplying by scalars and adding can get us the whole set. We can take the basis $\{x, x^2\}$.
Note that there are two elements in the basis, as there should be, since the $$\dim(\ker)=\dim(\text{domain})-\dim(\text{image})=3-1=2,$$
since the image $\left\{\left[\begin{matrix}a\\a\end{matrix}\right] : a\in \mathbb{R}\right\}$ is of dimension 1.
A: For a polynomial $p(x)=ax^2+bx+c$, $p(0)=c$. The nullspace of $T$ is all polynomials such that $T(p)=\begin{bmatrix}
        p(0) \\
        p(0) \\
        \end{bmatrix}=\begin{bmatrix}
        0 \\
        0 \\
        \end{bmatrix}$
Therefore the nullspace is all polynomials $p(x)=ax^2+bx$.
A: Recall the definition of kernel. Let $V,W$ be vector spaces over the same field of scalars and let $T:V\to W$ be a linear map. The kernel of $T$ denoted by $\ker T$ is the set of all $v \in V$ such that $T(v) = 0$.
So, let $T:P_2(\Bbb R)\to \Bbb R^2$ be the map you defined, i.e.: $T(p) = (p(0),p(0))$. One arbitrary element of $P_2(\Bbb R)$ is of the form $p(x)=ax^2+bx+c$. So we have $p(0)=c$. In that case, we have the following:
$$T(ax^2+bx+c)=(c,c).$$
So, what must be $ax^2+bx+c$ to give us $T(ax^2+bx+c)=(0,0)$? It must of course be such that $c=0$. So all elements of the kernel are of the form $ax^2+bx$. And so we found:
$$\ker T=\{p \in P_2(\Bbb R) : p(x)=ax^2+bx, \quad a,b\in \Bbb R\}$$
A: 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.
