Let $(V, (\cdot,\cdot))$ be a finite-dimensional inner product space, and let $H$ be a Hermitian operator on $V$. We know that $H$ is diagonalizable, and its eigenvalues are real.
We consider the following situation. Let $\{v_i\}$ be a basis for $V$, not necessarily orthogonal. Let $\{w_i\}\subset V$ be a dual basis, defined by the property $(w_i, v_j) = \delta_{ij}$. Define the matrix $A$, whose component is $A_{ij} = (w_i, Hv_j)$.
Question: Is $A$ diagonalizable? If so, can we say something on the eigenvalues of $A$ (for example, they are real)?
(This question is motivated from physics, and I have some physical expectation that the answer to the above question is true.)
$A$ is just the matrix representation of $H$ in the basis $\{v_1, \dots, v_n\}$. This is because $(w_i, Hv_j)$ is the $i$th component of $Hv_j$ in this basis. Evidently then, all your claims are true.