Trouble understanding the basics of a dual space and solving for its basis - Example from *Linear Algebra by Friedberg* This example comes from "Linear Algebra by Friedberg, et.al - pg. 120". It is example 4. The following is a screenshot of the example:

for reference the function $f_{i}$ is defined as $f_{i}(x) = a_{i}$, where
$$[x]_{\beta} = (a_{1}, a_{2}, \dots, a_{n})$$
is the coordinate vector of $x$ relative to the basis $\beta$.
The mechanics of the problem I understand, but it is the reasoning where I'm having an issue. I have the following questions:

*

*I could see why we would need to determine a formula for $f_{1}$ because up to now it is just an arbitrary component, but how we go about it is confusing. I see why $1 = f_{1}(2,1)$, it is because with respect to our basis $\beta$ it is the first component. But then why do we rewrite $(2,1)$ in terms of the standard basis vectors?


*I see how $f_{1}(e_{1}) = -1$ and $f_{1}(e_{2}) = 3$, but now using this idea are we then treating the vector $(x,y)$ as "scalars" when applied to $f_{1}$?
All of the same of course follows for $f_{2}$, but clarifying $f_{1}$ will take care of that. I think my main issue is with why rewriting the vector in terms of the standard basis vectors and how we are getting a formula from that.
 A: *

*Another way of finding $f_{1}$ and $f_{2}$ (without using the standard basis) would be to consider an arbitrary $(x, y) \in \mathbb{R}^{2}$ and writing
$$ (x, y) = f_{1}(x, y)(2, 1) + f_{2}(x, y)(3,1). $$
This gives us two equations for $f_{1}(x, y)$ and $f_{2}(x, y)$. After solving them, we get that $f_{1}(x, y) = 3y - x$ and $f_{2}(x, y) = x - 2y$.
Once we have a basis $\beta$ for $V$, the dual basis $\beta^{*}$ gives us the coordinates of any $v \in V$ with respect to $\beta$. Similarly, $\beta$ gives us the coordinates of any functional in $V^{*}$ with respect to $\beta^{*}$. To be more specific:
Let $\beta = (v_{1}, \ldots, v_{n})$ be a basis for $V$ and $\beta^{*} = (f_{1}, \ldots, f_{n})$ the corresponding dual basis.
Then, for any $v \in V$,
$$ v = f_{1}(v) v_{1} + \ldots + f_{n}(v) v_{n} .$$
Similarly, if $f \in V^{*}$,
$$ f = f(v_{1}) f_{1} + \ldots + f(v_{n}) f_{n} .$$


*It is a consequence of linearity:
$$ f_{1}(x, y) = f_{1}( x(1, 0) + y(0,1) ) = xf_{1}(1, 0) + yf_{1}(0,1). $$
A: Recall that $V^*$ is the space of linear functionals  i.e. linear functions from $V\to \mathbb{F}$ (in this case $\mathbb{F}=\mathbb{R}$) so that treats your second question.
Now note that by definition, we must have that $f_1(2, 1) = 1, f_1(3, 1)= 0$ because of how we ordered the basis. We want to explicitly determine what the function is. Now since it has to be $\textbf{linear}$, we want to write it in terms of its components because in terms of $e_1$, this will tell its $x$-component and likewise for $e_2$ for the $y$-component. Note that $f(c, 0)= cf(1, 0)$.
