Proof that the matrix of the dual map is the transpose

Proof

1. Write $B = (b_1, \dots, b_n)$ and $C = (c_1, \dots, c_m)$. Let $B^* = (b_1^*, \dots, b_n^*)$ and $C^* = (c_1^*, \dots, c_m^*)$ be the dual bases, so $b_l^*(b_j) = \delta_{lj}$ and $c_i^*(c_k) = \delta_{ik}$.
2. Here $\delta_{ij} = 1$ if $i = j$ and $\delta_{ij} = 0$ otherwise.
3. Let $[T]_{C,B} = (a_{ij})$, an $m \times n$ matrix. By the definition of the matrix of a linear map, $T(b_j) = \sum_{k=1}^{m} a_{kj} c_k$ for $1 \le j \le n$.
4. The dual map is $T^* \colon W^* \to V^*$, $T^*(f) = f \circ T$. Its matrix $[T^*]_{B^*, C^*} = (d_{li})$ is $n \times m$.
5. By the same definition, $T^*(c_i^*) = \sum_{l=1}^{n} d_{li} b_l^*$ for $1 \le i \le m$.
6. Fix $i \le m$ and $j \le n$. Evaluating step 5 at $b_j$ gives $T^*(c_i^*)(b_j) = \sum_{l=1}^{n} d_{li} \delta_{lj} = d_{ji}$.
7. On the other hand, by step 4 and step 3, $T^*(c_i^*)(b_j) = c_i^*(T(b_j)) = \sum_{k=1}^{m} a_{kj} c_i^*(c_k)$.
8. Since $c_i^*(c_k) = \delta_{ik}$, the sum in step 7 equals $a_{ij}$.
9. By steps 6 to 8, $d_{ji} = a_{ij}$ for all $i \le m$ and $j \le n$.
10. By the definition of the transpose, step 9 says $[T^*]_{B^*, C^*} = \left([T]_{C,B}\right)^T$.