6

Let $T$ be a transformation from $P_2$ to $P_2$ (where $P_2$ is the space of all polynomials with degree less than or equal to $2$)

$$T(f(t)) = f''(t)f(t)$$

I'm tempted to say that this is not a linear transformation because

$$T(f(t) + g(t)) = (f''(t) + g''(t))(f(t) + g(t))$$

Which does not equal

$$T(f(t)) + T(g(t))$$

But I'm not sure if I did that correctly...

Hagen von Eitzen
  • 374,180
Andy
  • 95
  • 5
    It's simpler to check the scalar multiplication condition. – hardmath Jul 06 '13 at 15:27
  • 3
    Basically, yes, it's correct, but although it looks obvious that in general $(f''(t)+g''(t))(f(t)+g(t)) \neq f''(t)f(t) + g''(t)g(t)$, give an explicit example to show the non-equality. (But what hardmath said.) – Daniel Fischer Jul 06 '13 at 15:27
  • Thanks for the help. I ended up showing it failed both scalar multiplication and addition properties. Thanks! – Andy Jul 06 '13 at 15:46

3 Answers3

9

$T(t^2+t)=2(t^2+t)\neq2(t^2)+0(t)=T(t^2)+T(t)$

Don't bother yourself, for showing that a property is not satisfied you only need to bring a counterexample. In fact $T(f+g)$ can be $T(f)+T(g)$ for some suitable $f$ and $g$. So saying that these are not equal for all $f$ and $g$ is a wrong sentence. For example if $f$ and $g$ be linear polynomials then both sides will be zero!

AmirHosein Sadeghimanesh
  • 3,865
6

You are correct, but to complete the job, you should provide a counterexample--that is, specific $f,g\in P_2$. such that $T(f+g)\ne T(f)+T(g)$. Alternately, find a scalar $\alpha$ and an $f\in P_2$ such that $T(\alpha f)\ne\alpha T(f)$.

Cameron Buie
  • 102,994
3

If $f(t)=at^2+bt+c$, then $T(f)(t)=2a^2t^2 +2abt+2ac$. The $a^2$ in the quadratic term looks suspicuously non-linear. In fact, if $f(t)=t^2$, then $T(f)(t) = 2t^2$ and $T(-f)(t)=2t^2$ as well. Assuming $2\ne0$, this contradicts linearity as that requires $T(f)+T(-f)=0$.

On the other hand, if $2=0$ in your base fiedl, then $T$ is the zero map and trivially linear.

Hagen von Eitzen
  • 374,180