Corollary 1

The notations being those of th. 1, suppose in addition that (C) is not satisfied, that $\Phi$ is non-degenerate and that, for every $x \in \mathbf{E}$, there exists $\rho \in \mathbf{A}$ such that $\Phi(x, x) = \rho \bar{\rho}$. Then $\mathbf{E}$ admits an orthonormal basis for $\Phi$.

So indeed let $(e_i)$ ($i = 1, \ldots, n$) be an orthogonal basis of $\mathbf{E}$. Put $\Phi(e_i, e_i) = \alpha_i$. We have $\alpha_i \neq 0$ for $i = 1, \ldots, n$ since $\Phi$ is non-degenerate. By assumption, there exist elements $\beta_i$ of $\mathbf{A}$ such that $\alpha_i = \beta_i \overline{\beta_i}$ for $i = 1, \ldots, n$; we have $\beta_i \neq 0$. On putting $f_i = \beta_i^{-1} e_i$, we have $\Phi(f_i, f_i) = \beta_i^{-1} \alpha_i \overline{\beta_i^{-1}} = 1$ for all $i$, and $\Phi(f_i, f_j) = 0$ for $i \neq j$. Thus $(f_i)$ is an orthonormal basis.