Proposition 1

If $f$ is a hermitian form on $E$, and $M$ a vector subspace of $E$ such that $f(x, x) = 0$ for all $x \in M$, then we also have $f(x, y) = 0$ for every pair of points $x, y$ in $M$.

Let $f$ be a hermitian form on $E$; the set $N$ of all $x \in E$ such that $f(x, y) = 0$ for all $y \in E$ is a vector subspace of $E$. It follows from (3) that, if $x_1 \equiv x_2$ (mod. $N$) and $y_1 \equiv y_2$ (mod. $N$), we have $f(x_1, y_1) = f(x_2, y_2)$; hence, on the quotient space $E/N$ we define a sesquilinear form $f$ by putting $f(\dot{x}, \dot{y}) = f(x, y)$ for all $x \in \dot{x}$ and all $y \in \dot{y}$; it is clear that $f$ is hermitian and that the relation «$f(\dot{x}, \dot{y}) = 0$ for all $\dot{y} \in E/N$» implies $\dot{x} = 0$ in $E/N$, in other words (A, IX) $f$ is separating. We say that $f$ is the separating hermitian form associated with $f$.