Question

In the vector space _2(\mathbb{R})" align="absmiddle" class="latex-formula"> consisting of real polynomials of a degree no more than 2, the following subset is considered:
$U=\left \{ \right.P(x)\in P_2(\mathbb{R})|P(4)=0\left. \right \}$

Determine a linear transformation $f: P_2(\mathbb{R}) \mapsto \mathbb{R}$ that has $U$ as a kernel.

Show and explain what you do along the way, step-by-step :-)
If you write the answer in hand, will you be kind to write as clearly as possible.
I would be very grateful for that and will appreciate the help you can give :-)

Answer 1

We could express any vector $p\in P_2(\mathbb R)$ as

$p=ax^2+bx+c$

So any vector $u\in U$ would have coefficients $a,b,c$ that satisfy

$16a+4b+c=0\implies c=-16a-4b$

from which we can show that any such vector is a linear combination of some other vectors:

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

More explicitly, we've shown that $u$ is a linear combination of the vectors $x^2-16$ and $x-4$; in other words, $u\in\mathrm{span}\{x^2-16,x-4\}$, and in fact these two vectors form a basis for $U$. But this set does not span all of $P_2(\mathbb R)$ because there's no combination of these two vectors that can be used to obtain a constant. We want the transformation to be usable for any vector in $P_2(\mathbb R)$, so we need to add an additional vector extend the basis. We can do this simply by appending $1$ into the spanning set. (Do check that the vectors remain linearly independent.)

Now we want the transformation to map those polynomials $p(x)$ for which $p(4)=0$ to the zero vector. We know which vectors belong to the basis of $U$, so we need

$f(x^2-16)=0$

$f(x-4)=0$

$f(1)=1$

where the choice of the assignment for $f(1)$ is arbitrary, so long as it's non-zero.