Kernel and image of linear operator - matrix representation Let $P_3(\mathbb{R})$ be a real vector space with the basis $V = (1,x,x^2)$. 
Find the kernel and the image of the image of the linear operator: 
$$D: P_3(\mathbb{R}) \to P_3(\mathbb{R})$$
$$p(x) \to p'(x)$$

With respect to the basis $V$ we know that: 
$$D(1) = 0 + 0x + 0x^2$$ 
$$ D(x) = 1 + 0x + 0x^2$$
$$ D(x^2) = 0 +2x + 0x^2$$ 
So the matrix representation would be: 
$$ _V[D]_V = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}$$
Thus we know that $x_1 =$ free variable, $x_2 = 0$ and $x_3 = 0$. The null space of the matrix representation is $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$
The image of the matrix representation is $(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix})$
My question is, how do I find the kernel and the image of the linear operator, when I know the Null space and the image of the corresponding matrix representation? 
 A: The matrix is just a representation of the linear operator with respect to some basis. In this case you took the 'standard' basis $\beta=\{1,X,X^2\}$ of $\mathbb{R}[X]_{\leq2}$, so we have $\ker(L)=\mathrm{span}\{1\}$ and $\mathrm{Im}(L)=\mathrm{span}\{1,X\}$, by transforming the kernel and image of the matrix representation back to the polynomial vector space. Formally this is done by an inverse coordinate transformation.
A: If i have understood your query correctly then you can use the fact that $\mathcal{P}_3(\mathbf{F})$ and $\mathbf{F}^3$ are isomorphic to each other denote $\mathcal{P}_3(\mathbf{F})\cong\mathbf{F}^3$ then assuming that you know that 
$$\operatorname{null}[D]_V = \operatorname{span}\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\right)\text{ }\text{ and }\text{ }\operatorname{range}[D]_V = \operatorname{span}\left(\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\2\\0\end{pmatrix}\right)$$
and the fact that $V = 1,x,x^2$ is the basis in question it follows that the corresponding spaces in $\mathcal{P_3}(\mathbf{F})$ are $\operatorname{null}D = \operatorname{span}(1)$ and $\operatorname{range}D = \operatorname{span}(1,2x)$.
