Question

In the vector space
`P_2(\mathbb{R})` 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.