General Question on Linear Functionals and the dual basis Suppose we are working in $X:=\mathbb{R}^2$ with basis vectors $e_1,e_2$ and we are trying to find the dual basis $\{f_1,f_2\}$ why are we able to say that for $x\in \mathbb{R}^2$ that $x = f_1(x)e_1+f_2(x)e_2$ and what does this necessarily accomplish? How does this help in determining what the dual basis is?
 A: If $f:V\to\mathbb{F}$ (You can take $\mathbb{F}$ to be any field. Take $\mathbb{R}$ if it makes you happy) be a Linear Functional. Then if
$\{v_{1}, v_{2}, \dots, v_{n}\}$ is a basis for $V$, the dual basis corresponding to $\{v_{1}, v_{2}, \dots, v_{n}\}$ is by definition the functionals $\{f_{1}, f_{2}, \dots, f_{n}\}$ such that $f_{i}(v_{j})=\delta_{i,j}$, where
$$
\delta_{i,j} = \begin{cases} 
1\,, i=j \\ 
0\,, i\neq j 
\end{cases}
$$
is the Kronecker delta
Take $V=\mathbb{R}^{n}$ .and take a ordered basis
$\{e_{1}, e_{2}, \dots, e_{n}\}$ By definition of dual basis, if
$x = \sum_{i=1}^{n} c_{i}e_{i}$ then
$f_{j}(x) = f_{j} \bigl( \sum_{i=1}^{n} c_{i}e_{i} \bigr) 
= \sum_{i=1}^{n} c_{i} f_{j}(e_{i}) = 
\sum_{i=1}^{n} c_{i} \delta_{i,j} = c_{j}$.
So $x = \sum_{j=1}^{n} f_{j}(x) e_{j}$.
Now to your question as to what it accomplishes is that it gives you the coefficients in terms of the dual basis. So if you directly define
$f_{j}(e_{i}) = \delta_{i,j}$ without knowing a priori what the definition of the dual basis is, then you end up with precisely what it should have been expected.
