Theorem 1

Suppose that $A$ is a field and $E$ a finite-dimensional vector space of dimension $n$ over $A$. Then, for every Hermitian form $\Phi$ on $E$, $E$ admits an orthogonal basis, unless the following conditions are simultaneously satisfied:

(C) $A$ is commutative of characteristic 2, the antiautomorphism $J$ is the identity, $\Phi$ is alternating and nonzero.

Let us reason by induction on $n$, the result being obvious for $n = 0$. We may suppose $\Phi \neq 0$. If (C) is not satisfied, lemma 1 shows that there exists an element $x \in E$ such that $\Phi(x, x) \neq 0$. Let $H$ be the subspace of $E$ orthogonal to $x$; it is of dimension $\geq n - 1$, and, since $x \in \mathbf{H}$, $\mathbf{H}$ is exactly of dimension $n - 1$. If the restriction $\Psi$ of $\Phi$ to $\mathbf{H}$ does not satisfy (C), there exists, by the induction hypothesis, a basis $(e_2, \ldots, e_n)$ of $\mathbf{H}$ which is orthogonal for $\Psi$; putting $e_1 = x$, we obtain an orthogonal basis $(e_1, e_2, \ldots, e_n)$ of $\mathbf{E}$. It remains to examine the case where $\mathbf{A}$ is a commutative field of characteristic 2, where $\mathbf{J}$ is the identity, and where $\Psi$ is alternating and $\neq 0$. There then exist $y, z$ in $\mathbf{H}$ such that $\Psi(y, z) \neq 0$; put $e_1 = x + y$; in order that $x + \lambda z (\lambda \in \mathbf{A})$ be orthogonal to $e_1$, it is necessary and sufficient that $0 = \Phi(x + y, x + \lambda z) = \Phi(x, x) + \lambda \Psi(y, z)$, a condition which determines $\lambda$ uniquely; the scalar $\lambda$ being thus chosen, we have $\Phi(x + \lambda z, x + \lambda z) = \Phi(x, x) \neq 0$, hence the restriction $\Psi'$ of $\Phi$ to the subspace $\mathbf{H}'$ of $\mathbf{E}$ orthogonal to $e_1$ is not alternating; we may consequently apply the induction hypothesis to $\mathbf{H}'$, which proves the theorem.

When (C) is satisfied, there is obviously no orthogonal basis for $\Phi$.