Proof of the dimension of the dual space by the dual basis

Proof

We show that the dual basis is a basis of $V^*$.

1. Let $n = \dim V$ and let $(b_1, \dots, b_n)$ be a basis of $V$. Write $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij} = 0$ if $i \ne j$.
2. For each $i$, the extension of values on a basis gives a linear form $b_i^* \colon V \to F$ with $b_i^*(b_j) = \delta_{ij}$ for all $j$. These form the dual basis $(b_1^*, \dots, b_n^*)$ in the dual space $V^*$.
3. The family $(b_1^*, \dots, b_n^*)$ is linearly independent. Suppose $\sum_{i=1}^n \lambda_i b_i^* = 0$ with $\lambda_i \in F$.
4. The operations of $V^*$ are pointwise. Evaluating at $b_j$ gives $0 = \sum_{i=1}^n \lambda_i b_i^*(b_j) = \sum_{i=1}^n \lambda_i \delta_{ij} = \lambda_j$. This holds for each $j$.
5. The family $(b_1^*, \dots, b_n^*)$ spans $V^*$. Let $g \in V^*$, and let $h = \sum_{i=1}^n g(b_i)\, b_i^*$.
6. For each $j$, $h(b_j) = \sum_{i=1}^n g(b_i) \delta_{ij} = g(b_j)$. So $g$ and $h$ are linear maps $V \to F$ that agree on a basis.
7. By the uniqueness part of the same result, $g = h$. So $g$ lies in the span of the $b_i^*$.
8. By steps 3 to 7, $(b_1^*, \dots, b_n^*)$ is a basis of $V^*$ with $n$ elements.
9. So $V^*$ is finite-dimensional, and its dimension is $n = \dim V$.