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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?

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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

1 Answer 1

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 $H(p(x))=p'(x)+p'(0)=0$.

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