[proofplan] For each $i$, define $\varepsilon_i$ on basis vectors by the Kronecker condition $\varepsilon_i(e_j) = \delta_{ij}$ and extend by linearity. Verify that the resulting functionals are linearly independent (by evaluating a dependence relation at basis vectors) and span $V^*$ (by expressing any functional as a linear combination using its values on the basis). [/proofplan] [step:Construct the dual basis elements by the Kronecker condition] For each $i \in \{1, \dots, n\}$, define $\varepsilon_i: V \to \mathbb{F}$ by specifying values on the basis: \begin{align*} \varepsilon_i(e_j) = \delta_{ij} = \begin{cases} 1 & \text{if } i = j, \\ 0 & \text{if } i \neq j, \end{cases} \end{align*} and extending linearly: $\varepsilon_i\bigl(\sum_{j=1}^n a_j e_j\bigr) = a_i$. By the [Universal Property of Linear Maps](/theorems/380), each $\varepsilon_i$ is the unique linear map with these prescribed values, so $\varepsilon_i \in V^*$. [/step] [step:Verify linear independence by evaluation at basis vectors] Suppose $\sum_{i=1}^n c_i \varepsilon_i = 0$ (the zero functional). Evaluating at $e_j$ for each $j$: \begin{align*} 0 = \Bigl(\sum_{i=1}^n c_i \varepsilon_i\Bigr)(e_j) = \sum_{i=1}^n c_i \delta_{ij} = c_j. \end{align*} So $c_j = 0$ for all $j$, proving $(\varepsilon_1, \dots, \varepsilon_n)$ is linearly independent. [/step] [step:Show the dual basis spans $V^*$] Let $\theta \in V^*$ be arbitrary. Define $c_i = \theta(e_i)$ for each $i$. For any $v = \sum_{j=1}^n a_j e_j \in V$: \begin{align*} \Bigl(\sum_{i=1}^n c_i \varepsilon_i\Bigr)(v) = \sum_{i=1}^n c_i a_i = \sum_{i=1}^n \theta(e_i)\, a_i = \theta\Bigl(\sum_{i=1}^n a_i e_i\Bigr) = \theta(v), \end{align*} where the penultimate equality uses linearity of $\theta$. Since this holds for all $v \in V$: $\theta = \sum_{i=1}^n c_i \varepsilon_i$. [/step] [step:Note uniqueness from the universal property] Each $\varepsilon_i$ is uniquely determined by the Kronecker condition and linearity, by the [Universal Property of Linear Maps](/theorems/380). Therefore the dual basis is unique. [/step]