The Wikipedia article for Hermitian matrices gives an alternative characterization for Hermitian matrices link. The statement I want to prove is:
$$ A^* = A \iff \langle w,Av \rangle_H = \langle Aw,v \rangle_H $$
This was proven for the standard inner product by a similar question.
I want to prove it for the general case with an arbitrary hermitian inner product on complex vector spaces.
I tried it using
$$ \langle w,Av \rangle_H = w^*HAv $$
with $H$ being a hermitian matrix and $ * $ marks the hermitian conjugate of a vector/matrix.
I don't really know how to show it without using that H and A commute.
I am thankful for any help!
Edit:
Note that I am using a positive definite hermitian form as the definition for the inner product. A hermitian form is defined as a sesquilinear form with $\langle w,v \rangle = \overline{\langle v,w \rangle}$.
Then there exists a hermitian matrix H being the transformation matrix of the hermitian form with
$$ \langle w,v \rangle_H = w^*Hv $$
Best Answer
The difference between this proof and the one you linked is mostly symbolic. We define the adjoint of $A$ denoted by $A^\dagger$ by $\langle A^\dagger i|j\rangle=\langle i|Aj\rangle$.
For any inner product, if $A=A^\dagger$ then $$A_{ij}=\langle i|Aj\rangle=\langle Ai|j\rangle=\langle j|Ai\rangle^*=A_{ji}^*$$ so $A$ is Hermitian. The converse follows by going in the opposite direction as well.