Is the map $T:V\rightarrow V$ defined by $T(p(x))=p'(x)$ injective, surjective, or both? Here $V=P(R)$
Let $p(x)=ax^2+bx+c$, so $p'(x)=2ax+b$.
I think it's not injective because to let $p'(x)=0$, $c$ can be any number.
I also think it's surjective because for any $q\in p'(x)$, it can be expressed by $a$ and $b$ in $p(x).$
Could someone correct my thinking?
 A: You should have a coherent definition of $V$ to proceed properly.
From the context, I'll assume that $V=\Bbb R_2[x]$. That is, the collection of polynomials of degree $\leq 2$. 
Your map $T:V\to V$ is defined by $T(p)=p^\prime$. There are lots of ways to determine the surjectivity of $T$. Perhaps the quickest is to note that $\deg(T(p))<2$ for every $p\in V$. Hence $T$ cannot be surjective.
A deeper analysis can be done by representing $T$ as a matrix. Note that $\beta=\{1,x,x^2\}$ is a basis for $V$ and that 
\begin{array}{rcccrcrcrcrcrcrc}
T(1) & = & 0      & = & \color{red}{0}\cdot 1 & + & \color{red}{0}\cdot x & + & \color{red}{0}\cdot x^2\\
T(x) & = & 1      & = & \color{blue}{1}\cdot 1 & + & \color{blue}{0}\cdot x & + & \color{blue}{0}\cdot x^2\\
T(x^2) & = & 2\,x & = & \color{green}{0}\cdot 1 & + & \color{green}{2}\cdot x & + & \color{green}{0}\cdot x^2
\end{array}
This implies that the matrix of $T$ relative to $\beta$ is
$$
[T]_\beta=
\begin{bmatrix}
\color{red}{0} & \color{blue}{1} & \color{green}{0} \\
\color{red}{0} & \color{blue}{0} & \color{green}{2} \\
\color{red}{0} & \color{blue}{0} & \color{green}{0}
\end{bmatrix}
$$
This matrix is rank two. The rank of $[T]_\beta$ is the dimension of the image of $T$. Since $\dim V=3$, this proves that $T$ is not surjective.
I should note that if $V=\Bbb R[x]$, that is if $V$ is the collection of all polynomials of arbitrary degree, then our map $T$ is surjective. Indeed, if $P$ is an antiderivative of $p$, then $T(P)=p$.
