From my notes${}^{\zeta}$, I have that,
So in short, I don't understand why, when we take the Hermitian conjugate of an outer product, say, $\lvert A_i \rangle\langle A_i \rvert$ it seems like we should just end up with an inner product, say, $\langle A_i \lvert A_i \rangle$ since I am just applying the result of eqn $(3.30)$ above, which sends $\lvert A_i \rangle$ to $\langle A_i \rvert$, and vice versa, which is tantamount to switching the order of them.
So, a more careful analysis is required...
I don't understand how eqn $(4.15)$ follows from eqn $(4.14)$, specifically, I know that for Hermitian operators $\hat{A}^{\dagger}=\hat{A}^*=\hat A$, since the eigenvalues of Hermitian operators are real. I understand this because the Hermitian adjoint, $\dagger$ operation means to transpose first, $\intercal$, then take the complex conjugate. But since there is only one operator, $\hat{A}$, we must have the case that $\hat{A}^{\intercal}=\hat{A}$.
In an attempt to understand this I will use 2 different indices, $i,j$, (as the fact that the same index was being used in my notes may be the source of my confusion). Starting with $(4.14)$, but only focussing on the outer product part, $$\lvert A_i \rangle\langle A_j \rvert$$ So if I can demonstrate that $\lvert A_i \rangle\langle A_j \rvert$ is invariant under $\dagger$, then I will have understood it.
I start by taking the Hermitian conjugate $$\Bigl(\lvert A_i \rangle\langle A_j \rvert\Bigl)^{\dagger}\stackrel{\intercal}{=}\Bigl(\lvert A_i \rangle\langle A_j \rvert\Bigl)^{\intercal}=\langle A_j \lvert A_i \rangle\stackrel{*}=\color{blue}{\Bigl(\langle A_j \lvert A_i \rangle\Bigl)^{*}=\langle A_i \lvert A_j \rangle}\ne \lvert A_i \rangle\langle A_j \rvert$$
In the blue part above I am making use of eqn $(3.27)$, which essentially states that, $\Bigl(\langle \beta \mid \gamma\rangle\Bigl)^*=\langle \gamma \mid \beta\rangle$ for Hilbert space vectors $\beta, \gamma$.
So in short, I am asking why, under the Hermitian adjoint operation, eqn $(4.14)$, $$\hat {A}=\sum_i A_i\lvert A_i\rangle\langle A_i \rvert$$ doesn't become $$\sum_i {A_i}^* \langle A_i \mid A_i \rangle?$$
${}^{\zeta}$ Sources from ICL dept. of physics