# 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?

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*}

In all of the contexts I've seen it being used, you have

$$\delta_{ij} = \begin{cases} 1, & \text{ for } i = j \\ 0, & \text{ for } i \neq j \end{cases}\tag{1}\label{eq1A}$$

The $$f_i$$ are linear functionals defined on $$V$$, with values in $$\Bbb F$$. The Kronecker Delta is defined by $$\delta_{ij}=\begin{cases}1,i=j\\0,i\ne j\end{cases}$$