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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?

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3 Answers 3

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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*}

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

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

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

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