Prove that matrix is Hermitian iff it's equal to its adjoint with general hermitian inner products

Question

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 $$ \langle w,v \rangle_H = w^*Hv $$

Answer 1

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.