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I'm having trouble understanding this question and the proper way to solve it. I don't understand the solution given and why this was the right way to answer it.


Problem:

For the vector space $P_3$ of polynomials of degree less than or equal to 3, let $T:P_3 \rightarrow R$ be the function

$$T(p)=p(2)+p(3)$$

Show that $T$ is a linear transformation, and find numbers a, b, and c so that

$$T(x+a)=T(x^2+b)=T(x^3+c)=0$$


Solution:

\begin{align*} T(p+q) &= (p+q)(2)+(p+q)(3) \\ &= \left(p(2)+q(2)\right)+\left(p(3)+q(3)\right)\\ &= \left(p(2)+p(3)\right)+\left(q(2)+q(3)\right)\\ &= T(p)+T(q) \\\\ T(cp) &= (cp)(2) +(cp)(3) \\ &= c\left(p(2)\right) + c\left(p(3)\right) \\ &= c\left(p(2)+p(3)\right) \\ &= cT(p) \end{align*} Because of these proofs $T$ is a linear transformation. \begin{align*} T(x+a)&=(2+a)+(3+a)\\ &=2a+5\\ &=0 \\\\ a&=-\frac{5}{2}\\\\ &...\\\\ b&=-\frac{13}{2}\\\\ c&=-\frac{35}{2} \end{align*}


My Confusion:

  • What is "$T:P_3\rightarrow R$" in words?
  • What is $P_3$ and how does it relate to the problem?
  • What does the lowercase $p$ in $T(p)=p(2)+p(3)$ represent and why can you replace it?
  • How does $\left(p(2)+p(3)\right)+\left(q(2)+q(3)\right)= T(p)+T(q)$? or $c\left(p(2)+p(3)\right)= cT(p)$?
  • Why isn't it intuitive that $T(p)=p(2)+p(3)$ is a transformation?
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Answers:

When $\;V_{\Bbb F}\;$ is a vector space over a field $\;\Bbb F\;$ , a linear transformation $\;T:V\to \Bbb F\;$ is called A linear functional .

$\;P_n\;$ is the vector space of all polynomials of degree less than or equal $\;n\;$ with coefficients from some field. Its dimension is $\;n+1\;$

$\;p\;$ represents a polynomial in $\;P_3\;$ .

The next question has an obvious answer if you've understood so far what's going on here.

I've no idea why you think such a thing is/would be "obvious", "intuitive" or whatever. This is mathematics and stuff must be proved. Period.

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