Is the following function a linear transformation

Question

I have the following exercise

L : $\mathbf P_3 \to \mathbf P_2$ $ f \to 2f` + (f(3))t^2$

What I have tried:

Let $f(s) = 3-s$ for an arbitrary $s$ and $a = 2 \in \mathbf R$.

Hence,

LHS: L($af$) = L($2f$) = ( $4f`$ ) + ($2f(3)t^2$) = 4 because f(3) = 3-3 = 0;

RHS: $a$L($f$) = 2L($f$) = 2( $2f`$ + $f(3)t^2$) = 2 (2 + 0 ) = 4.

$\to$ LHS = RHS

From what I've done so far I concluded that $L$ respects the second condition of a linear transformation, but I am not sure if this is a legitimate proof. Also when it comes to the first axiom, i.e: $L(x+y) = L(x) + L(y)$ I'm confused what to pick up as arguments since my set is a set of functions. Should it be $LHS: L(f+g) = 2(f+g)` + ( (f+g)(3)t^2) $?

Notations:

$\mathbf P_n = (t^0 = 1, t^1, t^2 .... t^n )$

$t^i(s) = s^i$

Answer 1

You have to prove that

• For all $a\in \mathbb{R}$ and for all polynomial $f\in P_3$, you have $L(af) = aL(f)$
• For all polynomials $f\in P_3$ and $g\in P_3$, you have $L(f+g) = L(f) + L(g)$

In what you have done, you have checked the fisrt condition only for very special cases ($f(s) = 1-s$ and $a=2$), but you really need to it for all of them

The easiest method is: take a polynomial $f$ of degree at most 3, and a real $a$, and compute $L(af)$ and $aL(f)$ as you did in the special cases, but with more abstract values : so start by writing $f(s) = b_0 + b_1s + b_2s^2 +b_3s^3$, and plug this in both expression. From there you should be able to finish the computations.

The second condition is nothing more difficult than the first, so if you understood the method for the first, you can do it all by yourself