Given the inner product $\langle x, y \rangle = y^HMx, \, x,y\in\mathbb{C}^n$ where $y^H = \bar{y}^T$ is the conjugate transpose and $M^H=M$ positive definite, I find the adjoint $A^*$ of $A\in\mathbb{C}^{n\times n}$ using:
$$\langle Ax,y \rangle = y^HMAx = y^H(A^*)^HMx = (A^*y)^HMx = \langle x, A^*y\rangle.$$
Then $MA = (A^*)^HM$, and thus $A^*=M^{-1}A^HM$ and $A=M^{-1}(A^*)^HM$. As far as I understand it, this can be interpreted as $A^H$ being defined wrt the canonical basis $\mathcal{E}=\{e_i :\, e_{ij}=\delta^i_j, \, i,j \in\{1,\ldots,n\}\}$, while its counterpart is really just $A^H$ defined wrt the basis formed by the columns of $M$: $\mathcal{B} = \{M_1, \ldots, M_n\}$. The same holds for $(A^*)^H$ and A. Could you recommend a reference discussing and illustrating the geometric interpretation more intuitively? Potentially with extension to infinite-dimensional vector spaces. I am guessing $M$ there would be some positive definite hermitian kernel.
How is the above reconciled with seeing $M$ as the metric tensor (i.e. $M_{ij} = \langle f_i, f_j \rangle$). I took the columns of $M$ to be the basis vectors, but what would be the geometric meaning of $f_i$ and $f_j$ in the above?