Proposition1. Let $x,y$ be linearly independent. For any vector $V$, there are uniquely determined scalars $\alpha$ and $\beta$ such that $V=\alpha x+\beta y$.
Addition. Let $v_{1},v_{2}$ be vectors with coordinate $\left( \begin{matrix} \alpha _{1}\\ \beta _{1}\end{matrix} \right)$, $\left( \begin{matrix} \alpha _{2}\\ \beta _{2}\end{matrix} \right)$ respectively. Then, $v_{1}+v_{2}$ being a vector, also has coordinate, say $\left( \begin{matrix} \alpha _{3}\\ \beta _{3}\end{matrix} \right)$. We want to expresss $\left( \begin{matrix} \alpha _{3}\\ \beta _{3}\end{matrix} \right)$ in terms of $\left( \begin{matrix} \alpha _{1}\\ \beta _{1}\end{matrix} \right)$, $\left( \begin{matrix} \alpha _{2}\\ \beta _{2}\end{matrix} \right)$.
Let's write down what we know:
$\left( \begin{matrix} \alpha _{1}\\ \beta _{1}\end{matrix} \right)$ is representation of $v_{1}$, that is, to say, $v_{1}=\alpha _{1}x+\beta _{1} y$.
$\left( \begin{matrix} \alpha _{2}\\ \beta _{2}\end{matrix} \right)$ is bla bla... $v_{2}$.... bla bla..., $v_{2}=\alpha _{2}x+\beta _{2} y$.
So, $v_{1}+v_{2}=(\alpha _{1}+\alpha _{2})x+(\beta _{1}+\beta _{2})y.$
Thus, $\left( \begin{matrix} \alpha _{3}\\ \beta _{3}\end{matrix} \right)=\left( \begin{matrix} \alpha _{1}+\alpha _{2}\\ \beta _{1}+\beta _{2}\end{matrix} \right)$.
Scalar Multiplication:
$\alpha \left( \begin{matrix} \beta \\ \gamma \end{matrix} \right) =\left( \begin{matrix} \alpha \beta \\ \alpha \gamma \end{matrix} \right)$. So, how can I prove this?