Is the map $T:P(\Bbb R)\rightarrow P(\Bbb R)$ defined by $T(p(x))=xp'(x)$ injective, surjective, or both? 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. 
 A: 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.
A: 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.
