Don't understand the notation conjugate transpose on a vector in assignment solution Here is my assignment question:

and here is the solution by my tutor:


My problem is that:
(1) If $x=-\frac{a_{ij}}{a_{ii}}e_i+e_j$, then how to apply conjugate transpose on $x$?
My rough work is:
$x^{*}=-\frac{a_{ij}}{a_{ii}}e_i^{*}+e_j^{*}$, but I do now know if I should take conjugate transpose on the coefficient $-\frac{a_{ij}}{a_{ii}}$ as well?
(2) Would you please work it out how to operate $x^{*}Ax$ step by step?
Any helps would be appreciated! Thanks so much.
 A: First,
$$
x^*=-\frac{\overline{a}_{ij}}{a_{ii}}e_i^*+e_j^*,
$$
because $a_{ij}$ is generally complex (since $A$ is Hermitian, diagonal entries are real). This follows simply from the definition: $x^*=\overline{x}^T$ and the fact that for complex $\alpha$ and $\beta$, $\overline{\alpha\beta}=\overline{\alpha}\overline{\beta}$. In addition, $e_i$'s are real (it actually does not matter whether we write $e_i^*$ or $e_i^T$).
Then
$$
\begin{split}
0<x^*Ax&=\left(-\frac{\overline{a}_{ij}}{a_{ii}}e_i^*+e_j^*\right)A\left(-\frac{a_{ij}}{a_{ii}}e_i+e_j\right)\\
&=
\frac{a_{ij}\overline{a}_{ij}}{a_{ii}^2}e_i^*Ae_i
-\frac{\overline{a}_{ij}}{a_{ii}}e_i^*Ae_j
-\frac{a_{ij}}{a_{ii}}e_j^*Ae_i
+e_j^*Ae_j\\
&=
\frac{a_{ij}\overline{a}_{ij}}{a_{ii}^2}a_{ii}
-\frac{\overline{a}_{ij}}{a_{ii}}a_{ij}
-\frac{a_{ij}}{a_{ii}}a_{ji}
+a_{jj}\\
&=\frac{a_{ij}\overline{a}_{ij}}{a_{ii}}
-\frac{\overline{a}_{ij}}{a_{ii}}a_{ij}
-\frac{a_{ij}}{a_{ii}}\overline{a}_{ij}
+a_{jj}\\
&=-\frac{a_{ij}\overline{a}_{ij}}{a_{ii}}
+a_{jj},
\end{split}
$$
where we used the fact that $e_i^*Ae_j=a_{ij}$ and $a_{ji}=\overline{a}_{ij}$.
Hence $a_{ij}\overline{a}_{ij}<a_{ii}a_{jj}$. The statement follows by using $|a_{ij}|^2=a_{ij}\overline{a}_{ij}$.
