Corollary 1

a) Let $x_1, \ldots, x_n$ be elements of the hilbertian space E. We have

where equality holds only if one of the $x_i$ is null, or if the sequence $(x_1, \ldots, x_n)$ is orthogonal.

b) Let $x_1, \ldots, x_n, y_1, \ldots, y_n$ be elements of the hilbertian space E. We have

if the vectors $x_i$ and $y_i$ are non-null; equality holds in (36) if and only if $(x_1, ..., x_n)$ and $(y_1, ..., y_n)$ are each an orthogonal basis of the same vector subspace of E.

The inequality (35) follows from prop. 6 by induction on $n$; the inequality (36) can be deduced by applying the Cauchy-Schwarz inequality in $\Lambda^n(E)$ and the formula (26) of V, p. 33.

Suppose that the sequence $(x_1, ..., x_n)$ is orthogonal. Then

since $\langle x_i | x_j \rangle = 0$ for $i \neq j$.

Now suppose that the vectors $x_1, ..., x_n$ are not null and do not form an orthogonal sequence. Since $\| x_1 \wedge ... \wedge x_n \|$ depends symmetrically on the vectors $x_1, ..., x_n$, we can assume that $x_1$ is not orthogonal to the subspace F of E generated by $x_2, ..., x_n$, and that F is not simply 0. We can decompose $x_1$ in the form $x'_1 + y$ with $y \neq 0$ in F and $x'_1$ orthogonal to F; then $\| x'_1 \| < \| x_1 \|$. But

and so

Suppose that the vectors $x_i$ and the vectors $y_i$ are not null. The equality in relation (36) is equivalent to the conjunction of the equalities

(37)

(38)

By the first part of the proof, the equalities (38) imply that each of the sequences $(x_1, ..., x_n)$ and $(y_1, ..., y_n)$ is orthogonal, which in turn implies that $x_1 \wedge ... \wedge x_n \neq 0$ and that $y_1 \wedge ... \wedge y_n \neq 0$. Under these conditions, relation (37) implies that there exists a scalar $\lambda \neq 0$ such that $y_1 \wedge ... \wedge y_n = \lambda x_1 \wedge ... \wedge x_n$ (V, p. 3, Remark 1); in other words, that $(x_1, ..., x_n)$ and $(y_1, ..., y_n)$ are bases of the same vector subspace of E (A, III, § 11, No. 13).