Let $P_3(\mathbb{C})$ be the complex vectorspace of complex polynomials of degree 2 or less. Let $\alpha,\beta\in\mathbb{C}, \alpha\neq\beta$.
Consider the function $L:P_3(\mathbb{C})\rightarrow\mathbb{C}^2$ defined as $L(p)=\begin{pmatrix} p(\alpha)\\ p(\beta) \end{pmatrix}$, for $p\in P_3(\mathbb{C})$.
Let $L$ be a linear transformation with base $V=(1,X,X^2)$ for $P_3(\mathbb{C})$ and the standardbase $\varepsilon =(e_1,e_2)$ for $\mathbb{C}^2$.
And let $R(\varepsilon [L]v)=\mathbb{C}^2$
1) Find the zero space $N(\varepsilon [L]v)$
Here is what I tried:
I found $\varepsilon [L]v$ to be $\varepsilon [L]v=\begin{pmatrix} 1\ \alpha\ \alpha^2 \\ 1\ \beta \ \beta^2\end{pmatrix}=A$.
Let $p(n)=a_0+b_0n+c_0n^2\in P_3(\mathbb{C})$. We can rewrite that to vector:
$p(n)=\begin{pmatrix} a_0\\ b_0 \\ c_0 \end{pmatrix}$.
Now suppose $\begin{pmatrix} a_0\\ b_0 \\ c_0 \end{pmatrix}\in N(\varepsilon [L]v)$, then we have
$A\begin{pmatrix} a_0\\ b_0 \\ c_0 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}$
=> $\begin{pmatrix} 1\ \alpha\ \alpha^2 \\ 1\ \beta \ \beta^2\end{pmatrix}\begin{pmatrix} a_0\\ b_0 \\ c_0 \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}$
=> $a_0+\alpha b_0+\alpha^2c_0=0$ and $a_0+\beta b_0+\beta^2c_0=0$.
We can rewrite the two equations above to
$(\alpha-\beta)b_0+(\alpha^2-\beta^2)c_0=0$. And using the fact that $\alpha\neq\beta$ we derieve that
$b_0=-(\alpha+\beta)c_0$
Which means that we have
$\begin{pmatrix} a_0\\ b_0 \\ c_0 \end{pmatrix}=\begin{pmatrix} a_0\\ -(\alpha+\beta)c_0 \\ c_0 \end{pmatrix}$
We can then conclude that
$$N(\varepsilon [L]v)=\{\begin{pmatrix} a_0\\ b_0 \\ c_0 \end{pmatrix}|a_0,b_0,c_0\in\mathbb{C}\}=\{\begin{pmatrix} a_0\\ -(\alpha+\beta)c_0 \\ c_0 \end{pmatrix}|a_0,c_0\in\mathbb{C}\}=\{a_0\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}+c_0\begin{pmatrix} 0\\ -(\alpha+\beta) \\ 1 \end{pmatrix}|a_0,b_0,c_0\in\mathbb{C}\}$$
It's almost completely correct, but an equation was lost when deduced $p(\alpha) = p(\beta)$ from $p(\alpha) =0$ and $p(\beta) =0$, namely that this common value is $0$, so that e.g. the constant $1$ polynomial is not in the nullspace. The missing equation will relate $a_0$ to $c_0$.
You can always make a quick verification of the dimensions, using the dimension theorem. Now $L$ maps surjectively $P_3(\Bbb C)$ to $\Bbb C^2$, so the nullspace will have dimension $3-2=1$.
Note also that it's not necessary to write up coordinates and matrix of $L$, we are just looking for those polynomials $f$ that simultaneously satisfy $f(\alpha) =0$ and $f(\beta) =0$.