Respuesta :
Yes. A trasformation [tex] T:V\to W [/tex] is linear if:
- [tex] T(\lambda v) = \lambda T(v) \ \forall \lambda \in \mathbb{R},\ v \in V[/tex]
- [tex] T(v+w) = T(v)+T(w)\ \forall v,w \in V[/tex]
The (second) derivative satisfies both these request: in fact, a generic element in p3 is [tex] f(x) = ax^3+bx^2+cx+d [/tex]. Let's prove that, for any real number [tex] \lambda [/tex], [tex] d_2(\lambda f) = \lambda d_2(f) [/tex].
We have:
[tex] d_2(\lambda f) = (\lambda f)'' = (\lambda(ax^3+bx^2+cx+d))'' = (\lambda ax^3+\lambda bx^2+\lambda cx+\lambda d)'' = 6\lambda ax + 2\lambda b [/tex]
On the other hand,
[tex] \lambda d_2(f) = \lambda f'' = \lambda(ax^3+bx^2+cx+d)'' = \lambda (6ax + 2b) = 6\lambda ax + 2\lambda bx [/tex]
The proof of the second point is the same: write the derivative of the sum, and the sum of the derivaties. You'll see that they are the same polynomial.
