Proof of the dimension of the annihilator by a dual basis

Proof

1. Let $n = \dim V$. By the result on the dimension of a subspace, $W$ is finite-dimensional. Let $k = \dim W$.
2. Let $b_1, \dots, b_k$ be a basis of $W$. It is linearly independent in $V$, so it extends to a basis of $V$.
3. By the definition of dimension, this basis of $V$ has $n$ elements. Write it as $b_1, \dots, b_n$.
4. Let $b_1^*, \dots, b_n^* \in V^*$ be the dual basis, so $b_i^*(b_j) = \delta_{ij}$. Here $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij} = 0$ otherwise.
5. We claim that $b_{k+1}^*, \dots, b_n^*$ is a basis of the annihilator $W^0 = \set{f \in V^* : f(w) = 0 \text{ for all } w \in W}$.
6. Each $b_i^*$ with $i > k$ lies in $W^0$. Indeed, each $w \in W$ is $w = \sum_{j=1}^{k} c_j b_j$ with $c_j \in F$. Then $b_i^*(w) = \sum_{j=1}^{k} c_j \delta_{ij} = 0$, since $j \le k < i$.
7. The forms $b_{k+1}^*, \dots, b_n^*$ are linearly independent. Suppose $\sum_{i=k+1}^{n} c_i b_i^* = 0$ with $c_i \in F$.
8. Evaluate the sum of step 7 at $b_j$ for $j > k$. This gives $\sum_{i=k+1}^{n} c_i \delta_{ij} = c_j = 0$. So all $c_j$ are $0$.
9. The forms $b_{k+1}^*, \dots, b_n^*$ span $W^0$. Let $f \in W^0$ and put $g = \sum_{i=k+1}^{n} f(b_i) b_i^*$.
10. For $j \le k$, we have $g(b_j) = 0$ by step 6, and $f(b_j) = 0$ because $b_j \in W$.
11. For $j > k$, we have $g(b_j) = \sum_{i=k+1}^{n} f(b_i) \delta_{ij} = f(b_j)$.
12. So $f$ and $g$ agree on the basis $b_1, \dots, b_n$. A linear map is determined by its values on a basis, so $f = g$.
13. By steps 6 to 12, $b_{k+1}^*, \dots, b_n^*$ is a basis of $W^0$. So $\dim W^0 = n - k$.
14. Hence $\dim W + \dim W^0 = k + (n - k) = n = \dim V$.