Cfrgtkky

Find a linear transformation $T:P_2(mathbb R)tomathbb R$, $P_2(mathbb R)$ being the vector space of polynomials of degree no more than $2$, such that $ker T=U$ where $U={P(x)in P_2(mathbb R)mid P(4)=0}$.

A vector $pin P_2(mathbb R)$ could be written as $p=ax^2+bx+c$, so any vector $uin U$ could then be expressed as

$$u=ax^2+bx+(-16a-4b)=a(x^2-16)+b(x-4)$$

Then $uinmathrm{span}{x^2-16,x-4}$. Where do I go from here to find a possible $T$?

I know that both of these vectors are linearly independent. I've seen similar questions/examples that suggest I "extend this set of vectors to a basis of insert vector space". Here, I imagine it means I need to add another vector to the spanning set so that it forms a basis of $P_2(mathbb R)$, which I believe would be satisfied by $1$ because no linear combination of the vectors in the spanning set can yield a constant.

The aforementioned examples went to say that any possible transformation $T$ should satisfy

$$begin{cases}T(x^2-16)=0\T(x-4)=0\T(1)neq0end{cases}$$

and I see why this should be the case, but I don't know what $T$ is supposed to look like, as I'm only really used to working with transformations $T:mathbb R^mtomathbb R^n$. Would it necessarily be a matrix? and of what dimensions?

More seriously, your base for $ker(T)$ could be simplified, but whatever.

Complete this family into a base of $mathbb R_2[X]$, by adding the constant polynomial $1$.

You then just have to define $T$ by stating : $T(1)=1$, $T(X-4)=0$ and $T(X^2-16)=0$.

As for every polynomial $P=aX^2+bX+c$, you have
$$P=a(X^2-16)+b(X-4)+c+16a+4b$$
you can check that $T(P)=P(4)$. Bingo !