finding the Ker(T) Let $T:\mathcal P_2(\Bbb R)\rightarrow \Bbb R^2$ be the linear transformation defined by 
$T (p(x))=[p(0),p(1)]$
Find a basis for Ker(T)?
I actually know how to find ker(T) once it becomes the standard matrix, I am just confused about the following: I was thinking that firstly, take any polynominal saying $p(x)=x^2+bx+c$, so $p(0)=c, p(1)=1+b+c$, then $T (p(x))=[c,1+b+c]$, but then I get stucked?
 A: Note that $T$ is surjective since for every $(x,y) \in \Bbb{R}^2$ for the polynomial $p(t) = (y-x)t+x$ we have
$$Tp= (p(0), p(1))=(x,y).$$
Therefore by the rank-nullity theorem we have
$$3 = \dim \mathcal{P}_2(\Bbb{R}) = \dim\ker T + \dim \operatorname{Im} T = \dim\ker T + \dim \Bbb{R}^2 = \dim\ker T + 2$$
so $\dim\ker T$ is one-dimensional.
Hence for the basis we only need one polynomial $p$ of degree $\le 2$ such that $p(0)=p(1) = 0$. The simplest example is
$$p(t) = t^2-t$$
so a basis for $\ker T$ is $\{p\}$.
A: The kernel is the following subset of $\mathcal{P}_2(\mathbb{R})$:
$$\{p \in \mathcal{P}_2(\mathbb{R}): T(p) = [0\; 0]\}.$$
Thus, $p \in \ker T \iff p(0) = p(1) = 0$.

In other words, the kernel consists precisely of those degree $\le 2$ polynomials which have $0$ and $1$ as roots. This means that the polynomial must be of the form
$$a(x - 0)(x - 1) = a(x^2 - x).$$
Thus, a basis for the kernel is $\{x^2 - x\}.$
A: Let $p(x)=ax^2+bx+c$, then 
$$T(p(x))=\begin{bmatrix}c \\ a+b+c\end{bmatrix}.$$
If $p(x) \in \ker(T)$, then $T(p(x))=\begin{bmatrix}0 \\ 0\end{bmatrix}$. So we get
\begin{align*}
a+b+c & = 0\\
c& =0.
\end{align*}
This gives us $a=-b$ and $c=0$. Thus 
$$\ker(T)=\{a(x^2-x) \, | \, a \in \mathbb{R}\}.$$
A basis for $\ker(T)$ is $\{x^2-x\}$.
