Proving Linearity of Transformation: f as a Real Number

[bifodus]

Let f: R --> R and let T: P2 --> F, and T(p) = p(f). Prove that T is a linear transformation.

P2 is the set of polynomials of degree 2 or less, and F is the set of all functions.

It seems to me that I can treat f as really just a real number, in which case it's no different from proving T(p) = p(x) for all x in R. Is it this simple?

Thanks

 

[Hurkyl]

You can treat f similarly to a real number, because you can add and multiply real-valued functions, but you can't treat it as a real number.

Secondly, you're not trying to prove T(p) = p(f): that's the definition of T. You're trying to prove that T is a linear transformation from P2 to F.

Finally, F is is not just a set, it's a vector space.

 

[bannadonna]

To prove that T(p) is a linear transformation, you must show that it has the properties of a linear transformation, that is:
(a) T(a+b)=T(a)+T(b) for a,b in P2
(b) T(ca)=cT(a) for a in P2 and c in R

 

[bifodus]

So let p1 and p2 be elements of P2
Then T(p1 + p2) = (p1 + p2)(f) = p1(f) + p2(f) = T(p1) + T(p2)
and T(cp1) = cp1(f) = cT(p1)

Is this sufficient to show that T is a linear transformation or am I leaving something out?

 

[bannadonna]

That looks right. Good job!