Is it true that if $T_A(x) = Ax$ then $T^*_A(x) = A^*x$? I tried to prove this for the standard inner product $$ \newcommand{\innp}[2]{\left\langle #1,#2 \right\rangle} \innp{Ax}{x} = x^tA^t\overline{x} = x^t\overline{A^*}\overline{x} = \innp{x}{A^*x} $$
but I can't seem to prove for a general inner product. Is it true? How do I generalize this for any inner product?
Note that any inner product on $\def\C{\mathbf C}\C^n$ has the form $$ \def\<#1>{\left<#1\right>}\<x,y>_B = x^t B\bar y $$ for some hermitian positive $B$. ($B$ has the entries $b_{ij} = \<e_i, e_j>_B$). To compute the matrix of the adjoint of multiplication by $B$, we can argue as follows: \begin{align*} \<Ax, y>_B &= x^tA^t B\bar y\\ &= x^t \overline{\bar A^t B^t} \bar y \\ &= x^t B\overline{\bar B^{-1}\bar A^t B^t y}\\ &= \<x, B^{-t}\bar A^t B^t y> \end{align*} So the matrix of $T_A^*$ (adjoint with respect to $\<,>_B$) for the standard basis is $B^{-t}\bar A^t B^t$.