Find a polynomial $p$ in $\Bbb P_2$ that spans the kernel of T, and describe the range of T.

Question

Define $ T: \Bbb P_2 \to \Bbb R^2 $ by $ T(p) = \begin{bmatrix} p(0) \\ p(1) \end{bmatrix} $.

For instance, if $ p(t) = 3 + 5t + 7t^2 $, then $ T(p) = \begin{bmatrix} 3 \\ 15 \end{bmatrix} $.

Find a polynomial $p$ in $\Bbb P_2$ that spans the kernel of T, and describe the range of T.

The answer given was:

Any quadratic polynomial $q$ for which $q(0)=0$ and $q(1)=1$ will be in the kernel of T. The polynomial $q$ must then be a multiple of $p(t) = t(t-1)$. Given any vector $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ in $\Bbb R^2$, the polynomial $p=x_1+(x_2-x_1)t$ has $p(0)=x_1$ and $p(1)=x_2$. Thus the range of T is all of $\Bbb R^2$.

I do not see how $p$ is a multiple of $t(t-1)$ but it does have a resemblance. How does $p$ span the kernel of T? What is the kernel of T?

Answer 1

For the quadratic polynomial $q$ to belong to $\ker T$, you must have $q(0) = 0$ and $\mathbf{q(1) = 0}$ (unlike what you quote). If you write $q(t) = a_0 + a_1 t + a_2 t^2$ then $q(0) = 0$ implies that $a_0 = 0$ and $q(1) = 0$ implies $a_1 + a_2 = 0$ so $q(t)$ has the form $q(t) = at - at^2 = at(t-1)$ for some $a \in \mathbb{R}$.