Let H be a Hilbert space and let $(e_n)_{n \in \mathbb{N}}$ be an orthonormal basis for H. Now define for each $T \in {\bf B}(H)$ the doubly infinite matrix $A = (\alpha_{nm})$ by setting $\alpha_{nm} = (Te_n|e_m)$. I am trying to find the matrix corresponding to $T^*$.
I know that since $T^*$ is the adjoint of T that for every $x,y \in H$, \begin{align*} (Tx|y) = (x|T^*y). \end{align*} Furthermore, since $(e_n)$ forms a basis for H, it suffices to show that \begin{align*} (Te_n|e_m) = (e_n|T^*e_m). \end{align*} So I expressed $Te_n = \alpha_{1n}e_1 + \cdots$ and observed that since $(e_n)$ is an orthonormal that \begin{align*} (Te_n|e_m) = \alpha_{mn}(e_m|e_m) = \alpha_{mn} = (Te_m|e_n). \end{align*} However, I could not see how to relate this to $(e_n|T^*e_m)$.
Also, it seems there is almost a canonical way to defined the matrix corresponding to $T^*$, however, I am not sure how to do it.
Any help or direction would be appreciated. Thank you in advance.
Hint:
$$\langle T^*e_n, e_m\rangle = \langle e_n, Te_m\rangle = \overline{\langle Te_m, e_n\rangle } = \overline{\alpha_{mn}}$$