About the theorem of dual basis Theorem for dual basis: Let V be a finite dimensional vector space and $\beta={u_1,u_2,...,u_n}$ be an ordered basis for V. Then there exists a basis $\beta^*={f_1,...,f_n}$ of $V^*$ such that $f_i(u_j)=\delta_{ij}$
So what is $f_i(u_j)=\delta_{ij}$ in terms of significance in a linear functional, is that a coordinate function similar to a coordinate vector in a vector space?
 A: Each linear functional $f_{i}(u_{j})$ is the linear mapping from the vector space $V$ onto its underlying field $\textbf{F}$ which associates $u_{i}$ to 1 and $u_{j}$ to zero, whenever $j\neq i$.
Thus, for instance, if one considers $\mathcal{B} = \{(1,0,0),(0,1,0),(0,0,1)\}$ to be the standard basis for $\textbf{R}^{3}$, one has that $f_{1}((1,0,0)) = 1$, $f_{1}((0,1,0)) = f_{1}((0,0,1)) = 0$.
More precisely, $f_{1}(x,y,z) = x$. Similarly, one also has that $f_{2}(x,y,z) = y$ and $f_{3}(x,y,z) = z$.
BONUS
Also, it is related to the coordinates of $\alpha\in V$ in the basis $\mathcal{B} = \{\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\}$ according to
\begin{align*}
[\alpha]_{\mathcal{B}} = (f_{1}(\alpha),f_{2}(\alpha),\ldots,f_{n}(\alpha))
\end{align*}
Moreover, given a basis $\mathcal{B} = \{\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\}$ of $V$, it is also used to express the coordinates of the linear functional $f\in V^{*}$ through the relation
\begin{align*}
f = f(\alpha_{1})f_{1} + f(\alpha_{2})f_{2} + \ldots + f(\alpha_{n})f_{n}
\end{align*}
A: In all of the contexts I've seen it being used, you have
$$\delta_{ij} = \begin{cases}
1, & \text{ for } i = j \\
0,  & \text{ for } i \neq j
\end{cases}\tag{1}\label{eq1A}$$
A: The $f_i$ are linear functionals defined on $V$, with values in $\Bbb F$.  The Kronecker Delta is defined by $\delta_{ij}=\begin{cases}1,i=j\\0,i\ne j\end{cases}$
