For $N\in \mathbb N$, equip $\mathbb C^N$ with the inner product $\langle\mathbf x,\mathbf y\rangle_N := N^{-1}\sum_i \overline x_i y_i$. Let $A$ be an $N\times M$ complex matrix. As an linear operator from $\mathbb C^M$ to $\mathbb C^N$, what is the adjoint of $A$?
Now for $\mathbf x\in \mathbb C^M$ and $\mathbf y \in \mathbb C^N$, we have $$ \langle A \mathbf x, y\rangle_N = N^{-1} \sum_{i=1}^N \sum_{j=1}^M \overline A_{ij} \overline{\mathbf x}_j \mathbf y_i = N^{-1} \sum_j \overline{\mathbf x}_j (A^*\mathbf y)_j = MN^{-1} \langle\mathbf x, A^* \mathbf y\rangle_M, $$ suggesting that the adjoint of $A$ as an operator is not the Hermitian transpose. I'm not sure how to interpret this.
I'm guessing that the matrix $A$ is not a representation of the linear transform with respect to an orthonormal basis?