I am working on proving the claim that $T$ is normal iff $[T]_{\beta}$ is normal with $\beta$ as an orthonormal basis. In my text it says follows immediately from Thm 6.10 which is just $[T^*]_{\beta} = [T]_{\beta}^*$ under $\beta$.
I'm running into an issue trying to prove the direction if $T$ is normal then $[T]_{\beta}$ is normal.
I've come to think that it suffices for me to show that the element $A_{ij} = B_{ij}$, where $A = [T]_{\beta}[T]_{\beta}^*$ and $B = [T]_{\beta}^*[T]_{\beta}$.
I know that an individual element of $[T]_{\beta}$ is of the form $[T]_{{\beta}_{ij}} = \langle T(v_j), v_i \rangle$. And analgously an element of $[T]_{\beta}^*$ is of the form $[T]_{{\beta}_{ij}}^* = \langle T^*(v_j), v_i \rangle$. Which I can extend to mean $\langle T^*(v_j), v_i \rangle = \overline{\langle T(v_i), v_j \rangle}$
Doing this an element $A_{ij}$ should be:
$$ A_{ij} = \sum_{k=1}^n \langle T(v_k), v_i \rangle_{ik} \overline{\langle T(v_j), v_k \rangle_{kj}} $$
This is taking into account what it means to multiply matrices. And for an element of $B_{ij}$:
$$B_{ij} = \sum_{k = 1}^n \langle T^*(v_k), v_i \rangle_{ik}\langle T(v_j), v_k \rangle_{kj}$$
I attempted manipulating both of these summations using all of the different variations of their inner products by taking conjugates and using the adjoint where needed, but I was not able to successfully get them to be equal. The closes I got by manipulating the summation of $B_{ij}$ was:
$$\langle T(v_i), v_k \rangle_{ik} \overline{\langle v_k, T(v_j) \rangle_{kj}}$$
but it didn't amount to anything in the end. Also I never used the fact that $T$ is normal in any of my work, so there must be something I'm doing wrong. A push in the right direction would be appreciated.