contestada

Let pn denote the vector space of polynomials in the variable x of degree n or less with real coefficients. Let d2:p3→p1 be the function that sends a polynomial to its second derivative. That is, d2(p(x))=p′′(x) for all polynomials p(x)∈p3. Is d2 a linear transformation?

Respuesta :

Yes. A trasformation [tex] T:V\to W [/tex] is linear if:

  1. [tex] T(\lambda v) = \lambda T(v) \ \forall \lambda \in \mathbb{R},\ v \in V[/tex]
  2. [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.

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