Proof that the matrix of the adjoint is the conjugate transpose

Proof

We compute both matrices entry by entry with the formula for coordinates in an orthonormal basis.

1. Let $V$ and $W$ be finite-dimensional inner product spaces over $\mathbb{K} = \R$ or $\C$, with inner products written $\ip{\cdot}{\cdot}$. Let $B = (b_1, \dots, b_n)$ and $C = (c_1, \dots, c_m)$ be orthonormal bases of $V$ and $W$.
2. Let $T \colon V \to W$ be linear, with adjoint $T^* \colon W \to V$. So $\ip{T(v)}{w} = \ip{v}{T^*(w)}$ for all $v \in V$ and $w \in W$.
3. By the definition of the matrix of a linear map, $[T]_{C,B} = (a_{ij})$ is the $m \times n$ matrix with $T(b_j) = \sum_{i=1}^{m} a_{ij} c_i$.
4. Likewise, $[T^*]_{B,C} = (d_{ji})$ is the $n \times m$ matrix with $T^*(c_i) = \sum_{j=1}^{n} d_{ji} b_j$, for $1 \le i \le m$.
5. By the formula for coordinates in the orthonormal basis $C$, the coordinates of $T(b_j)$ are $a_{ij} = \ip{T(b_j)}{c_i}$.
6. By the same formula in the orthonormal basis $B$, the coordinates of $T^*(c_i)$ are $d_{ji} = \ip{T^*(c_i)}{b_j}$.
7. By conjugate symmetry and step 2, $d_{ji} = \overline{\ip{b_j}{T^*(c_i)}} = \overline{\ip{T(b_j)}{c_i}} = \overline{a_{ij}}$.
8. By the definition of the conjugate transpose, the $(j, i)$ entry of $([T]_{C,B})^*$ is $\overline{a_{ij}}$.
9. So $[T^*]_{B,C}$ and $([T]_{C,B})^*$ have the same size $n \times m$ and the same entries. Hence $[T^*]_{B,C} = ([T]_{C,B})^*$.