Proposition 6

a) Let $\Psi$ be a hermitian form (resp. symmetric bilinear form if $\mathbf{A} = \mathbf{K}$ or $\mathbf{A} = \mathbf{K}(i)$) on $E$. There exists a basis of $E$ which is orthonormal for $\Phi$ and orthogonal for $\Psi$.

b) Suppose $\mathbf{A} = \mathbf{K}$ or $\mathbf{A} = \mathbf{K}(i)$, and let $\Psi$ be an alternating bilinear form on $E$. There exists a basis of $E$ which is orthonormal for $\Phi$ and with respect to which the matrix of $\Psi$ is of the form

where the $a_i$ are $\geqslant 0$ in K.

When $\Psi$ is hermitian, our assertion follows at once from Prop. 5 and the canonical correspondence between hermitian forms on $E$ and hermitian endomorphisms for $\Phi$. In the two other cases, let $u$ be the semilinear mapping of $E$ into itself defined at the beginning of this No. by the formula $\Psi(x, y) = \overline{\Phi(u(x), y)}$. Then $u^2$ is an $\mathbf{A}$-linear mapping; when $\Psi$ is symmetric (resp. alternating), one has $u = u^*$ (resp. $u = -u^*$), hence $u^2$ is hermitian; whence also

(resp. $-\Phi(u(x), u(x))$);

this shows that the hermitian form $(x, y) \to \Phi(u^2(x), y)$ is positive (resp. negative). Apply then Theorem 2, and let $V$ be a minimal element of the family of the subspaces $\neq \{0\}$ of $E$ stable for $u$ and for $u^*$. Since $u^2$ is a hermitian endomorphism, $V$ contains a proper vector $x \neq 0$ of $u^2$; put $u^2(x) = ax$, where $a \in \mathbf{K}$ (Prop. 5).

When $\Psi$ is symmetric, the inequality $\Phi(u^2(x), x) \geqslant 0$ shows that one has $a \geqslant 0$. Put $y = a^{1/2}x + u(x)$; one has $u(y) = a^{1/2}u(x) + ax = a^{1/2}y$, and $Ay$ is stable for $u$ and $u^* = u$. If $y = 0$, $Ax$ is stable for $u$ and $u^*$; in all cases, $V$ is a subspace of dimension 1, and our conclusion follows at once from Theorem 2.

When $\Psi$ is alternating, one has $a \leqslant 0$; put

One has $u(y) = -(-a)^{1/2}z, \ u(z) = (-a)^{1/2}y$ and

If $y = z = 0$, we have $a = 0$ and $u(x) = 0$, hence $Ax$ is stable under $u$ and $u^*$, $V$ is of dimension 1, and the matrix of the restriction of $\Psi$ to $V$ is zero since $\Psi$ is alternating. Otherwise, $y$ and $z$ are both $\neq 0$, they generate $V$, and since they are orthogonal, $V$ is of dimension 2 and the matrix of the restriction of $\Psi$ to $V$ is of the form $\begin{pmatrix} 0 & b \\ -b & 0 \end{pmatrix}$. Finally, by virtue of the formula $\Psi'(x', y') = \overline{\Phi(u(x'), y')}$, we have $\Psi'(x', y') = 0$ when $x'$ and $y'$ belong to two subspaces stable under $u$ and orthogonal for $\Phi$. This proves the existence of an orthonormal basis of $E$ (for $\Phi$) with respect to which the matrix of $\Psi$ has the indicated form. Q.E.D.

Exercises. — 1) Suppose that the hypotheses at the beginning of § 7 are satisfied; let $\Phi$ be a positive Hermitian form on $E$.

a) In order that $\Phi(x, x)\Phi(y, y) = \Phi(x, y)\overline{\Phi(x, y)}$, it is necessary and sufficient that $x$ and $y$ be linearly dependent or that the plane generated by $x$ and $y$ be isotropic.

b) Suppose that, for every $x \in E$, $\Phi(x, x)$ is a square in $K$. Now prove that for any two vectors $x, y$ of $E$, we have

If $\Phi$ is non-degenerate, the two members of this inequality can be equal only if $\alpha x + \beta y = 0$, where $\alpha$ and $\beta$ are two elements of $K$, not both zero and such that $\alpha \beta \leq 0$.

2) Suppose $A = K$ or $A = K(i)$. Let $X$ be a Hermitian square matrix of order $n$ over $A$, such that for every nonempty subset $H$ of $\{1, n\}$, we have $X_{H, H} \geq 0$ (notations of Proposition 3 of No. 1).

a) Let $\lambda$ be an element $> 0$ of $K$; now prove that the matrix $X + \lambda I$ is positive non-degenerate (use Proposition 3 of No. 1, reasoning by induction on $n$).

b) Deduce that the Hermitian matrix $X$ is positive.