This is a follow-up question to When does a symmetric algebra over a field of characteristic 0 fail to be semisimple?

Let $H$ be a symmetric algebra over $\mathbb{R}$ with symmetrizing trace $\tau:H\to \mathbb{R}$ such that the corresponding bilinear form $H \times H \to \mathbb{R}, (h,h') \mapsto \tau(hh')$ is positive definite. Assume also that $H$ is split over $\mathbb{R}$. Then we can choose an orthonormal basis $\mathcal{B}$ for the bilinear form. Thus
for each simple $H$-module, the Schur element $c_V$ is given by
$$
c_V = \frac{1}{\dim_\mathbb{R} V} \sum_{b \in \mathcal{B}}\chi_V(b)^2 \geq 0.
$$
Moreover, $c_V = 0$ implies that $\chi_V(h) = 0$ for all $h \in H$, but this contradicts the assumption $V \neq 0$.

Hence all the Schur elements are nonzero, so $H$ is semisimple.

Does anyone know a reference for this result?

Does anyone know of generalizations of this result? Specifically, is there a more general setting in which positive definiteness can be defined and can this still be used to conclude that $H$ is semisimple? For instance, suppose $H$ is an algebra over $\mathbb{R}(q)$, for an indeterminate $q$. Does the result hold without the assumption that $H$ is split?