Column space of complex matrix Let $A=\begin{pmatrix}
    1 & \alpha & \alpha^2\\
    1 & \beta  & \beta^2
  \end{pmatrix}, \alpha,\beta\in\mathbb{C},\alpha\ne\beta$. I know that the column space of A is supposed to be $\mathbb{C}^2$, but I'm not sure how to get there.
My attempt:
The column space of $A$ is the set of all vectors of the form
$c_1\begin{pmatrix}1\\1\end{pmatrix}+c_2\begin{pmatrix}\alpha\\\beta\end{pmatrix}+c_3\begin{pmatrix}\alpha^2\\\beta^2\end{pmatrix}=\begin{pmatrix}c_1+\alpha c_2+\alpha^2 c_3\\c_1+\beta c_2+\beta^2 c_3\end{pmatrix}$
Is this the correct way to go about it?

As $\alpha \neq \beta$, than the first two columns are linear independent, so the dimension of the image is at least $2$ that means your column space is $\mathbb{C}^2$.
If you like a constructive one more than make the following 
$$ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} + (-\alpha) \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} =  \begin{pmatrix} 0 \\ \beta - \alpha \end{pmatrix}$$
As $\alpha \neq \beta$ we can scale it to $$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$
Now we calculate $$\begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ \end{pmatrix}$$ 
and we have the canonical basis. The dimension argument would be enough anyway.
