0
$\begingroup$

Q: We're showing that this function is a linear transformation. Let $V$ be the set of polynomials in $x$ with real coefficients and suppose $L:V \to \mathbb{R}^2$ is defined by $L(p(x))=(p(1), p(2))$.

Then suppose $L:V \to \mathbb{R}^3$ is defined by $L(p(x))=(p(1), p(2), p(3))$.

So I've started this with $L(p(x))=(p(1), p(2))$ and want to show it holds under addition so would I take a function $q$ and have $((p+q)(1), (p+q)(2))$? Or do I need to do $(p(1)+p'(1), p(2)+p'(2))$? Am I on the right track with one of these?

  • 0
    What do you mean, exactly, by "it holds under addition"? And why would you consider the derivative $p'$? – 2017-02-13
  • 0
    For it to be a linear transformation it holds under addition and multiplication of scalars. And I didn't mean prime just like some other p – 2017-02-13

1 Answers 1

0

Hint: OK, so you need to show that if $p, q\in V$ and $c\in\mathbb{R}$, then $$L((p+q)(x)) = L(p(x))+L(q(x)),\quad L((cp)(x)) = cL(p(x)).$$ Now just write down the definitions of what those things mean in terms of the explicit definition for $L$ that you have, and the answer should pop out.

  • 0
    Just wondering here: what if I suppose p to be some random polynomial, say p=2x^2 + x. Then L(p)=(3,9). Could I show addition with: L(2x^2)+ L(x)? I.e. 2+1? – 2017-02-13
  • 0
    You need to show that this holds for *any* $p$ and $q$. In the particular case you gave, $L(2x^2) = (2\cdot 1^2, 2\cdot 2^2) = (2,8)$ and $L(x) = (2\cdot 1, 2\cdot 2) = (2,4)$, while $L(2x^2+x) = (2\cdot 1^2+1,2\cdot 2^2+2)$. And then $L(2x^2)+L(x) = L(2x^2+x)$. But again, you need to show this for arbitrary $p$ and $q$. – 2017-02-14