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

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

• Just to be clear: your definition of the inner product is $\langle x, y\rangle_H=x^* H y$? Jun 18 at 21:04
• @DavidRaveh yes exactly, I added it to the question. Jun 18 at 21:13
• Your definition of the inner product isn't valid, since it doesn't generally satisfy $\langle x,x \rangle_H\geq 0$ Jun 18 at 21:22
• @DavidRaveh you're right. I clarified it a bit. please see the edit in the question. Jun 18 at 21:46

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.
• I think I don't quite get it then. Isn't $\langle Ax, y\rangle_H = (Ax)^*y$ (copied from the other question) Implying that $H = I$ with $I$ being the identity matrix. Or is it that the basis can be chosen to fulfil $H=I$? Jun 18 at 22:17