Show that an hermitian operator is represented by an hermitian matrix. Let $\hat{A}$ be an hermitian operator.
I have to show that $\overline{A_{\alpha\beta}}=A_{\beta\alpha}$
So, if an operator is hermitian then $\hat{A} = \hat{A}*$
I started with this:
$\hat{A}*=\sum_{\alpha\beta}\overline{|u_{\alpha}\rangle A_{\alpha\beta}\langle u_{\beta}|}=\sum_{\alpha\beta}|u_{\beta}\rangle\overline{A_{\alpha\beta}}\langle u_{\alpha}|$
I don't know how to continue

Now note that $$\hat A=\sum_{\alpha\beta}|u_\beta\rangle A_{\beta\alpha}\langle u_\alpha|$$This must be equal to your final expression $\sum_{\alpha\beta}|u_{\beta}\rangle\overline{A_{\alpha\beta}}\langle u_{\alpha}|$ by Hermitianness of $\hat A$. Compare coefficients one by one. For instance by multiplying the two sums from the left by arbitrary basis bras $\langle u|$ and from the right by arbitrary basis kets $|u\rangle$.
