Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It has already been given that:

$$T\colon P_3\to P_4 \qquad T(p)=(x+2)p(x)$$

$$H\colon P_4\to P_3 \qquad H(p)=p'(x)+p'(0)$$

It is asked to show that $T$ is one-to-one but not onto, and that $H$ is onto but not one-to-one. How can I show that?

share|cite|improve this question
Which one are you having problems with ? what did you try ? – Belgi Sep 13 '12 at 12:54
A 1to1-criterion is to look at the Kernel, it's a well known fact that if the Kernel is {0} the application is 1to1. An onto-criterion is to look at dimension of the arrive space. Please, ask for more info. – Ivan Oct 27 '12 at 13:44
up vote 1 down vote accepted

Just use the definition to check that $T$ is one-to-one. To show that $T$ is not onto, note that for any $0\neq p(x)\in P_3$, $T(p(x))=(x+2)p(x)$ which is a polynomial having degree at least one. Therefore, for any degree zero polynomial $q(x)$, namely the constant, there does not exists any $p(x)\in P_3$ such that $T(p(x))=q(x)$.

Use the definition to check that $H$ is onto. To do it, you need to solve some linear equations. To show that $H$ is not one-to-one, just note that for any degree zero polynomials $p(x)$, we have $H(p(x))=p'(x)+p'(0)=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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