1
$\begingroup$

In my opinion, we can first write $p(x)$ as $ax^2+bx+c$. Then $xp'(x)=2ax^2+bx$. I think it is injective because in order for $2ax^2+bx=0$, $a, b, c$ in $ax^2+bx+c$ has to be zero, but I highly doubt my reasoning. I think it is not surjective because there is no $c$ in $2ax^2+bx$, and again, I doubt my reasoning. Someone helps would be great.

$\endgroup$
3
  • 2
    $\begingroup$ After looking at your example, should $T$ be $T(p(x)) := x p'(x)$ (NB the prime here but not in the question as written). $\endgroup$ Mar 1, 2016 at 14:06
  • $\begingroup$ Not all polynomials are of degree${}\leq2$. Also $x(ax^2+bx+c)=ax^3+bx^2+cx\neq 2ax^2+bx$. If you meant to ask a different question than you did, then please edit your title. (And it doesn't hurt spelling out the actual question (again) in the question.) $\endgroup$ Mar 1, 2016 at 14:10
  • $\begingroup$ Sorry, I just corrected the title $\endgroup$
    – CoolKid
    Mar 1, 2016 at 14:21

2 Answers 2

5
$\begingroup$

Let $p(x)=x$ and $q(x)=x+1$. Then $T(p(x))=x=T(q(x))$. So, $T$ is not 1-1.

Also, the range of $T$ can not include any non-zero constant polynomial. Therefore $T$ is not onto either.

$\endgroup$
2
  • $\begingroup$ Well, how can you be sure that $p(x)=x$ and $q(x)=x+1$? And I don't see $2ax^2+bx$ contain nonzero constant. $\endgroup$
    – CoolKid
    Mar 1, 2016 at 14:39
  • $\begingroup$ @CoolKid: I picked $p(x)$, $q(x)$ myself. It is an example. Also, you are right that $2ax^2+bx$ does not contain a non-zero constant. That's why you can never have a non-zero constant polynomial in the range of $T$. $\endgroup$ Mar 1, 2016 at 15:30
3
$\begingroup$

You're on the right track, but one should probably formalize the reasoning in both of the arguments. For example, as written, the proof of injectivity in the question only applies to polynomials of degree $\leq 2$.

If $T(p(x)) = 0$, we have $x p'(x) = 0$. Since the product of two nonzero polynomials is nonzero, this forces $p'(x) = 0$, and so the kernel of $T$ consists of the polynomials with zero derivative, namely, the constant polynomials. In particular, there is a nonzero polynomial $p$ such that $T(p(x)) = 0$, so $T$ is not injective.

From the above, we know that if $p$ is not in the kernel of $x$, then $\deg p > 0$ and $p'(x) \neq 0$, so $\deg T(p(x)) = \deg (x p'(x)) = \deg x + \deg p'(x) = 1 + \deg p - 1 = \deg p$. Thus, $T$ maps no polynomial to a polynomial of degree $0$, i.e., a constant polynomial, and hence $T$ is not surjective.

$\endgroup$
3
  • $\begingroup$ There is a problem with your argument. Your conclusion should be $p'(x)=0$, not $p(x)=0$. Therefore, $T(p(x))=0$ if and only if $p'(x)=0$, which proves that every constant polynomial maps to $0$ and therefore, $T$ is not 1-1. $\endgroup$ Mar 1, 2016 at 15:34
  • $\begingroup$ @IoannisSouldatos When I wrote this answer, the question posed asked about the transformation $T(p(x)) := x p(x)$, and not $T(p(x)) := x p'(x)$. In fact, after I wrote this answer, I read OP's examples more carefully, which lead me to ask in the comments under the question, whether they really meant the latter---apparently they did, as the question has now been updated. Unfortunately, someone else made a rather zealous edit to my answer that obscures this (and turns a correct answer to a previous version of the question into a wrong answer to the current version). $\endgroup$ Mar 1, 2016 at 16:36
  • $\begingroup$ I've updated my answer to address the revised version of the question. $\endgroup$ Mar 1, 2016 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.