TITLE: Is Hermitian matrix a self-adjoint operator w.r.t. just the standard inner product, or any inner product?
QUESTION [5 upvotes]: Given a linear map $L:V→V$ over an inner product space $(V,〈⟩)$, the adjoint of $L$, denoted by $L^*$, is a linear map $L^*:V→V$ s.t. $〈Lx,y⟩=〈x,L^* y⟩$ for any $x,y\in V$. $L$ is self adjoint if $L=L^*$ and then $〈Lx,y⟩=〈x,Ly⟩$.
Just consider $V=\Bbb C^n$. I am confused by Hermtian matrix. It is easy to prove a Hermitian matrix $H$ is self-adjoint under the standard inner product $〈x,y⟩=y^*x$, by $〈Hx,y⟩=y^*Hx=(H^*y)^*x=(Hy)^*x=〈x,Hy⟩$, but I am not sure if it is actually self adjoint for any inner product. If not, it would be great if someone could help give a counterexample.

REPLY [3 votes]: Unless $H$ is a scalar multiple of the identity matrix, the answer is always "no". Just take any positive definite matrix $A$ that doesn't commute with $H$ and define $\langle x,y\rangle=y^\ast Ax$. For $v\mapsto Hv$ to be self-adjoint with respect to this inner product, you need $y^\ast AHx=\langle Hx,y\rangle=\langle x,Hy\rangle=y^\ast HAx$ for any two vectors $x$ and $y$, but that would imply $AH=HA$, which by assumption is false.