Proposition 1

Suppose that $A$ is a commutative field. Let $\Phi$ be a hermitian form on $E$, and let $(x_n)$ ($n = 1, 2, \ldots$) be a sequence (finite or infinite) of linearly independent vectors of $E$ such that, for every $n$, the subspace $E_n = Ax_1 + \cdots + Ax_n$ is non-isotropic. Let $D_{jn}$ ($j \leq n$) be the cofactor of $\Phi(x_j, x_n)$ in the matrix $(\Phi(x_s, x_t))_{(s,t=1,\ldots,n)}$. Then $D_{nn} \neq 0$ for every $n$. Put

Then, for every $n$, $(e_1, \ldots, e_n)$ is an orthogonal basis of $E_n$ and we have

Indeed, since the restriction of $\Phi$ to $E_{n-1}$ is non-degenerate, one has $D_{nn} \neq 0$ (§ 2, prop. 3); note that $D_{11} = 1$ since the determinant of the empty matrix is equal to 1. Formulas (2) imply first that one has $e_n \equiv x_n$ (mod. $E_{n-1}$) for every $n$, and hence that the $e_n$ are linearly independent, and that $(e_1, \ldots, e_n)$ is a basis of $E_n$. For every $j < n$, one has

(chap. III, § 6, no. 1, formula (12)); hence $e_n$ is orthogonal to $E_{n-1}$, and in particular to $e_j$ for $j < n$. On the other hand, one has

(chap. III, § 6, no. 1, formula (10)). This proves our assertions.

With the notation of prop. 1, we say that the sequence $(e_n)$ is obtained from the sequence $(x_n)$ by the Gram-Schmidt orthogonalization process.