§ 7. Hermitian forms and ordered fields

Throughout this paragraph, let K denote a maximal ordered field (hence commutative and of characteristic zero; cf. chap. VI, § 2), and suppose that we are in one of the following three cases:

1°) A = K, J is the identity; 2°) A is the field K(i), obtained by adjunction to K of a square root i of −1, and, for every λ ∈ A, $\overline{\lambda}$ is the conjugate of λ (chap. II, § 7, no 7). 3°) A is the quaternion field over K corresponding to the pair (−1, −1) (or, as we shall say for brevity, the quaternion field over K), and, for every λ ∈ A, $\overline{\lambda}$ is the conjugate of λ (cf. chap. II, § 7, no 8 and chap. VIII, § 11, no 2).

If Φ is a hermitian form on E, we therefore have, in all cases, $\Phi(x, x) \in K$ for every $x \in E$ , since $\Phi(x, x) = \overline{\Phi(x, x)}$ .

1. Positive hermitian forms.

Definition 1

A hermitian form $\Phi$ on $E$ is said to be positive (resp. negative) if $\Phi(x, x) \geqslant 0$ (resp. $\Phi(x, x) \leqslant 0$ ) for every $x \in E$ .

When $A = K$ , one also says that the quadratic form

Q(x) = \frac{1}{2} \Phi(x, x)

associated with

\Phi

is positive (resp. negative).

Suppose $E$ is finite-dimensional over $A$ , and let $(e_i)$ ( $i = 1, \ldots, n$ ) be an orthogonal basis of $E$ ( $§ 6$ , no 1, th. 1). For a hermitian form $\Phi$ on $E$ to be positive, it is necessary and sufficient that $\Phi(e_i, e_i) \geqslant 0$ for $i = 1, \ldots, n$ . Let $\Phi$ be a positive non-degenerate hermitian form on $E$ ; since every positive element of $K$ is a square, hence of the form $\rho \overline{\rho}$ ( $\rho \in A$ ), there exist orthonormal bases for $\Phi$ in $E$ ( $§ 6$ , no 1, cor. 1 of th. 1).

Proposition 1

Suppose $E$ is finite-dimensional, and $A = K$ or $A = K(i)$ . If $\Phi$ is a positive non-degenerate hermitian form on $E$ , then the extensions of $\Phi$ to $\bigotimes^p E$ and $\wedge^p E$ ( $p > 0$ ), as well as the inverse form of $\Phi$ , are positive non-degenerate hermitian forms.

This follows at once from the existence of an orthonormal basis of $E$ , and from prop. 2, $§ 6$ , no 1.

Prop. 1 remains true if « positive non-degenerate » is replaced everywhere by « positive ».

Proposition 2

Let $\Phi$ be a positive hermitian form on $E$ . For $x, y$ in $E$ , one has

\Phi(x, y) \overline{\Phi(x, y)} \leqslant \Phi(x, x) \Phi(y, y).

The inequality is indeed immediate when the vectors $x$ and $y$ are proportional. We may therefore suppose that $x$ and $y$ are linearly independent. Let $A'$ be the (commutative) subfield $K(\Phi(x, y))$ of $A$ , $F$ the vector plane over $A'$ generated by $x$ and $y$ , and $\Phi_F$ the restriction of $\Phi$ to $F$ ; the latter takes its values in $A'$ . By Proposition 1, the discriminant of $\Phi_F$ with respect to the basis $(x, y)$ is $\geqslant 0$ . Now this discriminant is $\Phi(x, x)\Phi(y, y) - \Phi(x, y)\overline{\Phi(x, y)}$ . Q.E.D.

Corollary

The set of isotropic vectors of E is the orthogonal subspace $E^0$ of E for $\Phi$ . In order that $\Phi$ be non-degenerate, it is necessary and sufficient that $\Phi(x, x) > 0$ for every $x \neq 0$ .

Proposition 3

Suppose E finite-dimensional and A commutative ( $\mathbf{A} = \mathbf{K}$ or $\mathbf{A} = \mathbf{K}(i)$ ). Let $\Phi$ be a hermitian form on E, X its matrix with respect to a basis $(x_j)$ ( $j = 1, \ldots, n$ ) of E; for every subset H of $\{1, n\}$ , denote by $X_{H, H}$ the minor of X obtained by deleting the rows and columns with indices $j \notin H$ (Chap. III, § 6, No. 3).

a) *If $\Phi$ is positive non-degenerate, then $X_{H, H} > 0$ for every subset H of $\{1, n\}$ .

b) *Conversely, if, on putting $H_j = \{1, j\}$ , we have $X_{H_j, H_j} > 0$ for $j = 1, \ldots, n$ , then $\Phi$ is positive non-degenerate.

Suppose first that $\Phi$ is positive non-degenerate. The elements $(x_j)$ ( $j \in H$ ) form a basis of a subspace F of E, and the minor $X_{H, H}$ is the discriminant of the restriction $\Phi_F$ of $\Phi$ to F with respect to this basis; now, since $\Phi_F(x, x) > 0$ for every $x \neq 0$ in F, $\Phi_F$ is positive non-degenerate (Corollary to Proposition 2); hence $X_{H, H} > 0$ (Proposition 1). To prove b), let us note that, with the notation of Proposition 1, § 6, No. 1, the minor $X_{H_j, H_j}$ is equal to $D_{j+1, j+1}$ ; there therefore exists (Proposition 1, § 6, No. 1) an orthogonal basis $(e_j)$ ( $j = 1, \ldots, n$ ) of E such that $\Phi(e_j, e_j) > 0$ for $j = 1, \ldots, n$ ; consequently $\Phi$ is positive non-degenerate.

Remark

It follows from the existence of orthonormal bases that two non-degenerate positive hermitian forms on two vector spaces of the same finite dimension are equivalent (§ 1, no 6). Let then L be a finite-dimensional hermitian space over A, whose metric form is positive non-degenerate, and let $V_1$ and $V_2$ be two linear varieties of the same dimension in L (§ 6, no 6); since the restrictions of the metric form to the directions $T_1$ and $T_2$ of $V_1$ and $V_2$ on the one hand, and to the orthogonal subspaces $T_1^0$ and $T_2^0$ on the other hand, are equivalent, there exists a unitary automorphism $u$ of the space $T$ of the translations of $L$ such that $u(T_1) = T_2$ and $u(T_1^0) = T_2^0$ ; there therefore exists a displacement $v$ of $L$ such that $v(V_1) = V_2$ . Let $(a, b), (a', b')$ be two pairs of points of $L$ ; for there to exist a displacement $v$ of $L$ such that $v(a) = a'$ and $v(b) = b'$ , it is therefore necessary and sufficient (with the notation of § 6, no 6) that one has $e(a, b) = e(a', b')$ ; the element $\sqrt{e(a, b)}$ of $A$ (chap. VI, § 2, no 4) is called the distance between $a$ and $b$ in the hermitian space $L$ .

2. The law of inertia.

Theorem 1 (« law of inertia »)

Suppose that $A$ satisfies the hypotheses at the beginning of this paragraph, and that $E$ is of finite dimension $n$ . Let $\Phi$ be a hermitian form on $E$ . Then:

a) There exists a decomposition of $E$ into a direct sum of the subspace $E^0$ orthogonal to $E$ , and of two subspaces $E^+$ and $E^-$ such that the restriction of $\Phi$ to $E^+$ (resp. $E^-$ ) is positive (resp. negative) and non-degenerate.

b) There exists an orthogonal basis $(e_i)_{1 \leq i \leq n}$ of $E$ such that

\Phi\left( \sum_{i=1}^n \xi_i e_i, \sum_{i=1}^n \eta_i e_i \right) = \sum_{i=1}^s \xi_i \overline{\eta_i} - \sum_{i=s+1}^{s+t} \xi_i \overline{\eta_i}.

c) The dimensions $s$ of $E^+$ and $t$ of $E^-$ are the same for all the decompositions into a direct sum satisfying the conditions stated in a) ; the integer $s$ (resp. $t$ ) is the maximum of the dimensions of the subspaces $F$ of $E$ such that the restriction of $\Phi$ to $F$ is positive (resp. negative) and non-degenerate.

d) The rank of $\Phi$ is $s + t$ .

e) If $\Phi$ is non-degenerate, its index is equal to $\inf(s, t)$ (§ 4, no 2, def. 2).

Indeed, let $(x_i)$ ( $i = 1, \ldots, n$ ) be an orthogonal basis of $E$ (§ 6, No. 1, Theorem 1); recall that $\Phi(x, x) \in K$ for every $x \in E$ . We may suppose that $\Phi(x_i, x_i) > 0$ for $i = 1, \ldots, s$ , $\Phi(x_i, x_i) < 0$ for $i = s + 1, \ldots, s + t$ and $\Phi(x_i, x_i) = 0$ for $i = s + t + 1, \ldots, n$ . This proves a), since we take for $E^+$ (resp. $E^-$ ) the subspace generated by $x_1, \ldots, x_s$ (resp. by $x_{s+1}, \ldots, x_{s+t}$ ), and $E^0$ is then generated by $x_{s+t+1}, \ldots, x_n$ . We deduce b) from this by observing that $\Phi(x_i, x_i)$ is of the form $\rho \overline{\rho}$ (resp.

- $\rho \bar{\rho} )\ (\rho \in A^*)$ for $i = 1, \ldots, s$ (resp. $i = s + 1, \ldots, s + t$ ). To prove c), consider a subspace P of E such that the restriction of $\Phi$ to P is positive and non-degenerate; we then have $P \cap (E^- + E^0) = \{0\}$ , and the sum $P + E^- + E^0$ is therefore direct; we conclude that $\dim P \leq \dim E^+ = s$ , and this proves c). Assertion d) follows at once from a).

Suppose finally that $\Phi$ is non-degenerate, put $q = \inf (s, t)$ , and show that q is the index of $\Phi$ . With the notations of b), the vectors $e_i + e_{s+i}$ (resp. $e_i - e_{s+i}$ ) ( $i = 1, \ldots, q$ ) generate a totally isotropic subspace F (resp. $F'$ ). Since K is of characteristic 0, $F + F'$ is generated by $e_1, \ldots, e_q, e_{s+1}, \ldots, e_{s+q}$ and is therefore non-isotropic. The restriction of $\Phi$ to the subspace $H = (F + F')^0$ is then positive (or negative) and non-degenerate, and H contains therefore no isotropic vector $\neq 0$ . Since $F^0$ contains F and H, and since $\dim(F + H) = \operatorname{codim} F' = \operatorname{codim} F$ , we have $F^0 = F + H$ . Thus an isotropic vector z orthogonal to F is necessarily in F, because, in the direct sum $F + H$ , the component of z in H is isotropic. Consequently F is a maximal totally isotropic subspace, and this proves e).

Definition 2

With the notations of Theorem 1, the pair $(s, t)$ is called the signature of $\Phi$ .

3. Reduction of a form with respect to a positive hermitian form.

In this No. we shall suppose that E is of finite dimension $n$ over $A$ , and we shall denote by $\Phi$ a non-degenerate positive hermitian form on E.

Since the linear mappings of E into $E^*$ associated with $\Phi$ are bijective, whatever the elements $x, y, z$ of E with $y \neq 0$ , there exists one and only one element $t$ of E such that $\Phi(x, z) = \overline{\Phi(t, y)}$ . In particular, if $A = K$ or $A = K(i)$ and if $u$ is a semilinear mapping for J of E into itself (chap. II, App. I, No. 1), there exists, for every $x \in E$ , one and only one element $u^*(x)$ such that for every $y \in E$ , $\Phi(x, u(y)) = \overline{\Phi(u^*(x), y)}$ . It is immediately seen that $u^*$ is a semilinear mapping for J of E into itself; it is called the adjoint of $u$ .

Remark 1

When J is the identity, we recover the notion of adjoint of a homomorphism defined in § 1, No. 8.

Remark 2

Suppose still that $A = K$ or $A = K(i)$ . Let $E^J$ be the right A-module defined in § 1, No. 2, Def. 5. The form $\Phi$ is a bilinear form on $E \times E^J$ , $\Phi^J$ is a bilinear form on $E^J \times E$ , and $u$ is an A-linear mapping of E into $E^J$ . The adjoint of this homomorphism, in the sense of § 1, No. 8, is an A-linear mapping of E into $E^J$ ; it is immediately seen that this coincides with the mapping $u^*$ defined above.

An endomorphism $u$ of E is said to be normal (for $\Phi$ ) if $uu^* = u^*u$ .

Examples of normal endomorphisms: 1) the unitary automorphisms for $\Phi$ ( $\S$ 6, No. 2), which are characterized by the relation $u^{-1} = u^*$ ( $\S$ 1, No. 8, cor. of prop. 8); 2) the endomorphisms $u$ such that $u^* = u$ ; these are called hermitian endomorphisms.

For every hermitian endomorphism $u$ of E, put $\Phi_u(x, y) = \Phi(u(x), y)$ ; we have

\Phi_u(y, x) = \Phi(u(y), x) = \Phi(y, u(x)) = \overline{\Phi(u(x), y)} = \overline{\Phi_u(x, y)}.

which shows that

\Phi_u

is a hermitian form on E. Conversely let

\Psi

be a hermitian form on E; since the mapping

s_\Phi

of E into

E^*

associated with

\Phi

is bijective, there exists, for every

x \in E

, one and only one element

u(x)

of E such that

\Psi(x, y) = \Phi(u(x), y)

; it is easily verified that

u

is a hermitian endomorphism of E. Thus

u \to \Phi_u

is a canonical bijection of the set of hermitian endomorphisms of E onto the set of hermitian forms on E.

Suppose that $A = K$ or $A = K(i)$ ; given a bilinear form $\Psi$ on E, there exists, for every $x \in E$ , one and only one element $u(x)$ of E such that $\Psi(x, y) = \overline{\Phi(u(x), y)}$ ; it is immediately seen that $u$ is a semilinear mapping (for J) of E into itself. Since $\Phi(u(x), y) = \Phi(y, u(x)) = \overline{\Phi(u^*(y), x)}$ , we see that, for $\Psi$ to be symmetric (resp. alternating), it is necessary and sufficient that $u^* = u$ (resp. $u^* = -u$ ).

Theorem 2

Let S be a set of endomorphisms of E (resp. of semilinear mappings of E into E when A = K or A = K(i)) stable under the mapping u → u*. Then, if V is a subspace of E stable under S, its orthogonal V⁰ is stable under S. On the other hand E is the direct sum of subspaces stable under S, minimal in the set of subspaces ≠ {0} and stable under S, and pairwise orthogonal.

Indeed let V be a subspace of E stable under S; whatever x ∈ V⁰, y ∈ V and u ∈ S, we have u*(y) ∈ V, whence $\Phi(y, u(x)) = \Phi(u*(y), x) = 0$ (resp. $\Phi(y, u(x)) = \overline{\Phi(u^*(y), x)} = 0$ ), and consequently u(x) ∈ V⁰; thus V⁰ is stable under S, which proves our first assertion. For the second we shall proceed by induction on the dimension n of E, the case n = 0 being trivial. For n ≠ 0 there exists a subspace V ≠ {0} of E stable under S and minimal, for example a stable subspace ≠ {0} of minimal dimension. It then suffices to apply the induction hypothesis to V⁰, since the adjoint of the restriction of u to V⁰ (with respect to the restriction of $\Phi$ ) is identical with the restriction to V⁰ of the adjoint of u.

Corollary 1

Suppose that A = K or that A = K(i). Let B be a subalgebra of $\mathscr{L}_A(E)$ , stable under the mapping u → u*. Then E is a semisimple B-module, and is the direct sum of simple submodules pairwise orthogonal. The algebra B is semisimple.

Indeed, since every B-submodule V of E has a supplement, for example V⁰, the B-module V is semisimple (chap. VIII, § 3, No. 3, prop. 7). Since every B-submodule ≠ {0} and minimal in E is simple, E is the direct sum of simple submodules pairwise orthogonal. Finally B is a semisimple algebra, since it admits a semisimple and faithful module E whose contramodule is of finite type (chap. VIII, § 5, No. 1, prop. 3).

Corollary 2

The assumptions and notations being those of cor. 1, suppose in addition that the algebra B is commutative. Then all its elements are semisimple (normal) endomorphisms of E. When A = K (resp. A = K(i)), E is the direct sum of simple B-submodules pairwise orthogonal, which are vector spaces of dimension 1 or 2 (resp. 1) over A.

The first assertion follows from chap. VIII, § 9, No. 1, prop. 2, since B is semisimple. On the other hand every simple B-module is isomorphic to a B-module of the form B/m, where m is a maximal ideal of B, hence to a commutative field L of finite degree over A; when A = K (resp. A = K(i)), the field L is necessarily isomorphic to K or K(i) (resp. K(i)), since K(i) is algebraically closed (chap. VI, § 2, No. 6, th. 3).

Proposition 4

Let u be a normal endomorphism of E. When A is equal to K(i) or to the quaternion field over K, there exists an orthonormal basis (for Φ) of E formed by eigenvectors of u. When A = K, u is semisimple and E is direct sum of stable subspaces for u, pairwise orthogonal, and of dimension 1 or 2.

First examine the case where A is commutative (A = K or A = K(i)). Then the subalgebra B = A[u, u*] of $\mathscr{L}_A(E)$ is commutative since u is normal; it is stable under the mapping $\varphi \to \varphi^*$ by virtue of formulas (32) and (33) of § 1, No. 8. The assertion concerning the case A = K then follows immediately from cor. 2 of th. 2. When A = K(i), this corollary also shows that E is direct sum of vector subspaces $Ax_i$ ( $i = 1, \ldots, n$ ) of dimension 1, pairwise orthogonal and stable for u; if one puts $e_i = (\Phi(x_i, x_i))^{-1/2} x_i$ , $(e_i)$ is the orthonormal basis sought.

When A is the quaternion field over K, it will suffice similarly to prove, by virtue of th. 2, that every minimal element of the set of subspaces $\neq \{0\}$ of E stable for u and $u^*$ is of dimension 1. Now such a subspace V necessarily contains an eigenvector $x \neq 0$ of u (*), as is seen by observing that the quaternion field A contains K(i) as an algebraically closed subfield, and by restricting to K(i) the field of

(*) If E is a left vector space over a noncommutative field A, and u an endomorphism of E, one still says that a vector $x \neq 0$ of E is an eigenvector for u if there exists $a \in A$ such that $u(x) = ax$ ; the scalar a is then called an eigenvalue of u. One will note that, for every $b \neq 0$ in A, the vector $bx$ is an eigenvector for u, and that the corresponding eigenvalue is $bab^{-1}$ .

scalars of V. It remains therefore only to show that the eigenvector x of u is also an eigenvector of u*. Put $u(x) = ax$ ( $a \in A$ ); one then has $\Phi(u(x), x) = a\Phi(x, x) = \Phi(x, x)a = \Phi(x, \overline{a}x)$ since $\Phi(x, x)$ belongs to the center of A, and on the other hand $\Phi(u(x), x) = \Phi(x, u^*(x))$ ; it follows that one has $\Phi(x, u^*(x) - \overline{a}x) = 0$ , and one can therefore write $u^*(x) = \overline{a}x + z$ , where z is a vector orthogonal to x. One therefore has

\Phi(u^*(x), u^*(x)) = \overline{a}a\Phi(x, x) + \Phi(z, z) = \Phi(u(x), u(x)) + \Phi(z, z).

Now, since u is normal, one has

\begin{align*} \Phi(u(x), u(x)) &= \Phi(x, u^*u(x)) = \Phi(x, uu^*(x)) \\ &= \Phi(uu^*(x), x) = \Phi(u^*(x), u^*(x)). \end{align*}

Consequently one has $\Phi(z, z) = 0$ , which, by assumption, entails $z = 0$ and $u^*(x) = \overline{a}x$ . QED.

Remark

It follows from th. 2 that the eigenspaces relative to two distinct eigenvalues of u are orthogonal.

Proposition 5

Let u be a hermitian endomorphism of E. The eigenvalues of u belong to K, and there exists an orthonormal basis of E formed of eigenvectors of u.

When A is equal to K(i) or to the quaternion field over K, it suffices, by virtue of prop. 4, to prove the first assertion; now, if x is an eigenvector $\neq 0$ of u and a the corresponding eigenvalue, one has, by virtue of the assumption $u = u^*$ ,

a\Phi(x, x) = \Phi(u(x), x) = \Phi(x, u(x)) = \Phi(x, x)\overline{a} ;

as $\Phi(x, x)$ is a nonzero element of the center of A, it follows that $a = \overline{a}$ , hence $a \in K$ . Consequently a hermitian matrix has all its eigenvalues in K. Suppose now that A is equal to K. The matrix M of u with respect to an orthonormal basis of E is then symmetric; it is therefore a hermitian matrix if one considers it as a matrix over K(i). The first part of the proof then shows that this matrix has all its eigenvalues in K, and admits therefore, if $E \neq \{0\}$ , nonzero eigenvectors in E. It follows that every stable subspace for u and minimal is necessarily of dimension 1, and the conclusion follows immediately, as in prop. 4.

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

\begin{pmatrix} 0 & a_1 & 0 & 0 \ldots 0 \\ -a_1 & 0 & 0 & 0 \ldots 0 \\ 0 & 0 & 0 & a_2 \ldots 0 \\ 0 & 0 & -a_2 & 0 \ldots 0 \\ \cdots & \cdots & \cdots & \cdots \end{pmatrix}.

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

\Phi(u^2(x), x) = \overline{\Phi(x, u^2(x))} = \Phi(u^*(x), u(x)) = \Phi(u(x), u(x))

(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

y = (-a)^{1/2}x + u(x), \quad z = (-a)^{1/2}x - u(x).

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

\Phi(y, z) = -a\Phi(x, x) - \Phi(u(x), u(x)) = -\Phi(ax, x) + \Phi(u^2(x), x) = 0.

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

\sqrt{\Phi(x + y, x + y)} \leq \sqrt{\Phi(x, x)} + \sqrt{\Phi(y, y)}.

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.

Remark 3

One assumes that $A = K$ or $A = K(i)$ and that $E$ is finite-dimensional. Let $\Phi_1, \Phi_2$ be two positive hermitian forms on $E$ , and let $V = (\alpha_{ij}), W = (\beta_{ij})$ be the matrices of these forms with respect to the same basis $(e_i)$ of $E$ . Show that the hermitian form $\Phi$ whose matrix $(\gamma_{ij})$ with respect to $(e_i)$ is such that $\gamma_{ij} = \alpha_{ij}\beta_{ij}$ for every pair of indices, is positive; moreover, if $\Phi_1$ and $\Phi_2$ are non-degenerate, the same is true of $\Phi$ . (In the calculation of $\Phi(x, x)$ , express the $\alpha_{ij}$ in terms of the values $\Phi_1(c_i, c_i)$ for an orthogonal basis $(c_i)$ of $E$ relative to $\Phi_1$ .)

Remark 4

One assumes that $A = K$ or $A = K(i)$ . Let $R$ be a hermitian matrix of order $n$ over $A$ ; one says that $R$ has signature $(s, t)$ if the hermitian form on $A^n$ having $R$ as matrix with respect to the canonical basis, has signature $(s, t)$ . Let $\Delta_k$ denote the principal minor of $R$ obtained by deleting in $R$ the rows and columns of index $> k$ ; one assumes that $\Delta_{s+t} \neq 0$ and that, for no index $k < s + t$ , $\Delta_k$ and $\Delta_{k+1}$ are simultaneously zero (cf. § 6, exerc. 1 b)). Show that if $\Delta_k = 0$ for some $k < s + t$ , $\Delta_{k-1}$ and $\Delta_{k+1}$ have opposite signs (method of exerc. 1 a) of § 6) and that the number $s - t$ is equal to

\operatorname{sgn} \Delta_1 + \operatorname{sgn} (\Delta_1 \Delta_2) + \cdots + \operatorname{sgn} (\Delta_{s+t-1} \Delta_{s+t})

(use exerc. 2 of § 6).

Remark 5

One assumes that $A = K$ or $A = K(i)$ ; let $R, S$ be two hermitian matrices over $A$ , of signatures $(s, t)$ and $(s', t')$ respectively (exerc. 4); show that the matrix $R \otimes S$ is a hermitian matrix of signature $(ss' + tt', st' + s't)$ .

Remark 6

Let $K$ be a maximal ordered field, $L$ a simple algebra of finite rank over $K$ . If $(s, t)$ is the signature of the symmetric bilinear form $(x, y) \to \operatorname{Tr}_{L/K}(xy)$ on $L$ , now prove that one has: $s - t = m$ if $L$ is isomorphic to a matrix algebra of order $m$ over $K$ ; $s - t = 0$ if $L$ is isomorphic to a matrix algebra over the field $K(i)$ ; $s - t = -2m$ if $L$ is isomorphic to a matrix algebra of order $m$ over the quaternion field over $K$ .

Remark 7

We suppose that $A$ satisfies the conditions at the beginning of § 7 and that $E$ is of finite dimension $n$ . Let $\Phi$ be a non-degenerate positive hermitian form on $E$ . a) Now prove that every similitude for $\Phi$ can be written in one and only one way as the product of a homothety of ratio $\geqslant 0$ in $K$ , and of a unitary transformation. b) For every basis $(a_i)_{1 \leq i \leq n}$ of $E$ , show that there exists one and only one orthonormal basis $(e_i)_{1 \leq i \leq n}$ of $E$ satisfying the following conditions: 1° for every $m$ such that $1 \leq m \leq n$ , the subspace generated by $a_1, \ldots, a_m$ is identical with the subspace generated by $e_1, \ldots, e_m$ ; 2° one has $\Phi(a_i, e_i) \geq 0$ in $K$ , for every index $i$ (cf. § 6, no 1, prop. 1). c) Deduce from b) that for every invertible square matrix $M$ of order $n$ over $A$ , there exists one and only one pair of matrices $(L, U)$ of order $n$ , such that $U$ is a unitary matrix, that $L = (\lambda_{ij})$ has only zeros below its diagonal and diagonal terms $\lambda_{ii}$ belonging to $K$ and $> 0$ , and finally that $M = LU$ .

Remark 8

On suppose $A = K$ ; soit $L$ un espace hermitien de dimension finie sur $A$ , dont la forme métrique est positive non dégénérée. Soient $M, N$ deux parties de $L$ , $u$ une bijection de $M$ sur $N$ telle que l’on ait $e(u(a), u(b)) = e(a, b)$ quels que soient les points $a, b$ de $M$ . Montrer qu’il existe un déplacement dont la restriction à $M$ soit égale à $u$ (raisonner par récurrence sur la dimension de la variété linéaire affine engendrée par $M$ , comme dans le procédé d’orthogonalisation de Gram-Schmidt). La proposition s’étend-elle au cas où A est égal à K(i) ou au corps des quaternions sur K ?

Remark 9

On suppose remplies les conditions du n° 3, et en outre que A est le corps des quaternions sur K. Soit u un endomorphisme normal de E. Montrer que E est somme directe de sous-espaces F_k (1 \leq k \leq r), deux à deux orthogonaux, tels que dans chacun des F_k il existe une base orthonormale (e_{ik}) (1 \leq i \leq n_k) et que l’on ait u(e_{ik}) = \lambda_k e_{ik} pour 1 \leq i \leq n_k, deux \lambda_k d’indices distincts n’étant pas transformés l’un de l’autre par un automorphisme intérieur de A. En outre, si (F'_k) est une seconde décomposition de E ayant les mêmes propriétés, avec un système (\lambda'_k) de valeurs propres, on a F'_k = F_k (à une permutation près des indices) et \lambda'_k = \alpha_k \lambda_k \alpha_k^{-1}; l’ensemble des vecteurs propres de u pour la valeur propre \lambda_k est le sous-espace sur le commutant A_k de \lambda_k dans A, engendré par les e_{ik} (1 \leq i \leq n_k).

Remark 10

On suppose remplies les conditions du n° 3, et en outre que A est égal à K(i) ou au corps des quaternions sur K. Soit u un endomorphisme normal de E.

a) Montrer que pour qu’un sous-espace vectoriel F de E soit tel que u(F) \subset F, il faut et il suffit que F soit engendré par des vecteurs propres de u ; on a alors u(F^0) \subset F^0 et u^*(F) \subset F.

b) Pour que u soit unitaire (resp. pour que u^* = u, u^* = -u), il faut et il suffit que pour toute valeur propre \lambda de u, on ait \lambda \overline{\lambda} = 1 (resp. \overline{\lambda} = \lambda, \overline{\lambda} = -\lambda).

c) On dit qu’un endomorphisme hermitien u de E est positif (resp. positif et non dégénéré) si la forme hermitienne (x, y) \to \Phi(u(x), y) qui lui correspond canoniquement est positive (resp. positive et non dégénérée) ; il faut et il suffit pour cela que toutes les valeurs propres de u soient \geq 0 (resp. > 0).

d) Montrer que pour tout entier m > 0, il existe un endomorphisme normal v de E tel que v^m = u. Si u est hermitien positif, il existe un seul endomorphisme hermitien positif v tel que v^m = u, et il existe un polynôme f \in K[X] tel que v = f(u) ; ce dernier résultat est aussi valable lorsque A = K.

Remark 11

On suppose remplies les conditions du n° 3, et en outre que A = K. Soit u un endomorphisme normal de E.

a) Soit V un élément minimal de l’ensemble des sous-espaces de E non réduits à 0, stables pour u et u* ; montrer que si V est de dimension 2, la restriction de u à V est une similitude directe de multiplicateur > 0.

b) Montrer que tout sous-espace de E stable pour u est aussi stable pour u*. (Soit E_0 l’espace vectoriel obtenu à partir de E par extension à K(i) du corps des scalaires ; \Phi est la restriction à E d’une forme hermitienne positive non dégénérée sur E_0 et u la restriction à E d’un endomorphisme normal u_0 de E_0 (pour cette forme) ; en outre E est l’ensemble des x \in E_0 invariants par une bijection semi-linéaire involutive j de E_0, et on a u_0j = ju_0. Appliquer alors l’exerc. 10 a.).

Remark 12

On suppose remplies les conditions du n° 3, et en outre que A est égal à K(i) ou au corps des quaternions sur K. Montrer que pour tout endomorphisme $u$ de $E$ , il existe une base orthonormale de $E$ par rapport à laquelle la matrice de $u$ n’ait que des zéros au-dessous de la diagonale. (Procéder par récurrence sur la dimension de $E$ , en considérant un vecteur propre de $u$ .)

Remark 13

Soient $A$ un corps vérifiant les conditions du début du § 7, $E, F$ deux espaces vectoriels de dimension finie sur $A$ , $\Phi$ (resp. $\Psi$ ) une forme hermitienne positive non dégénérée sur $E$ (resp. $F$ ). Montrer que, pour toute application linéaire $u$ de $E$ dans $F$ , l’adjoint $u^*$ de $u$ (pour $\Phi$ et $\Psi$ ; cf. § 1, no 8) est tel que $u^*u$ et $uu^*$ soient des endomorphismes hermitiens positifs (exerc. 10 c)) de $E$ et $F$ respectivement.

Remark 14

On suppose remplies les conditions du no 3. Soient $u$ un endomorphisme de $E$ , $h_1, h_2$ les deux endomorphismes hermitiens positifs de $E$ , tels que l’on ait $h_1^2 = u^*u, h_2^2 = uu^*$ (exerc. 13 et 10 d)).

a) Montrer qu’il existe un endomorphisme unitaire $\nu$ tel que $u = \nu h_1 = h_2 \nu$ et en particulier que $h_1$ et $h_2$ sont semblables (remarquer que $u(0) = \overline{h_1}(0)$ et que si $V$ est le sous-espace orthogonal à $\overline{u}(0)$ , on a $\Phi(u(x), u(x)) = \Phi(h_1(x), h_1(x))$ pour tout $x \in V$ . Pour que $\nu$ soit déterminé de façon unique, il faut et il suffit que $u$ soit bijective. Pour qu’on puisse prendre $\nu$ permutable avec $h_1$ , il faut et il suffit que $u$ soit normal.

b) Deduce from a) that every square matrix $M$ of order $n$ over $A$ can be written $UDV$ , where $U$ and $V$ are unitary matrices and $D$ a diagonal matrix whose elements are $\geqslant 0$ , and whose squares are the eigenvalues of $MM^*$ .

Remark 15

Suppose that the conditions of No. 3 are satisfied. Show that every positive Hermitian matrix $H$ over $A$ can be written in the form $LL^*$ , where $L = (\lambda_{ij})$ has only zeros below its diagonal and diagonal terms belonging to $K$ and $\geqslant 0$ ; moreover $L$ is uniquely determined by these conditions when $H$ is invertible (cf. exerc. 10 d) and 7 c)).

Remark 16

Suppose that the conditions of No. 3 are satisfied and in addition that $A = K(i)$ . Let $u$ be an endomorphism of $E$ .

a) The set of eigenvalues of $u$ is contained in the set $U$ of the values of $\Phi(x, u(x))$ when $x$ runs through the set of elements of $E$ such that $\Phi(x, x) = 1$ .

b) A subset $C$ of $A = K(i)$ is said to be convex if, for every pair of elements $(\xi, \eta) \in C^2$ and every $\tau \in K$ such that $0 \leqslant \tau \leqslant 1$ one has $\tau \xi + (1-\tau)\eta \in C$ . Show that the set $U$ is convex. (Reduce to the case where $n = 2$ ; by writing $u$ in the form $\nu + i\omega$ , where $\nu$ and $\omega$ are hermitian, and by replacing if necessary $u$ by $\lambda u$ , where $\lambda \in A$ and $\lambda \overline{\lambda} = 1$ , show that everything reduces to proving the following property. Let $f(\xi_1, \xi_2) = \xi_1 \overline{\xi}_1 + \xi_2 \overline{\xi}_2$ , $g(\xi_1, \xi_2) = a \xi_1 \overline{\xi}_1 + b \xi_2 \overline{\xi}_2$ ( $a \in K, b \in K$ ), $h(\xi_1, \xi_2) = \alpha \xi_1 \overline{\xi}_1 + \beta \xi_1 \overline{\xi}_2 + \beta \xi_2 \overline{\xi}_1 + \gamma \xi_2 \overline{\xi}_2$ ( $\alpha \in K, \gamma \in K, \beta \in A$ ); let $(\eta_1, \eta_2) \in A^2, (\zeta_1, \zeta_2) \in A^2$ be such that $f(\eta_1, \eta_2) = f(\zeta_1, \zeta_2) = 1$ , $g(\eta_1, \eta_2) = g(\zeta_1, \zeta_2) = 1$ , $h(\eta_1, \eta_2) > 0$ , $h(\zeta_1, \zeta_2) < 0$ ; there then exists $(theta_1, theta_2) \in A^2$ such that $f(\theta_1, \theta_2) = 1$ , $g(\theta_1, \theta_2) = 1$ and $h(\theta_1, \theta_2) = 0$ . One will begin by observing that for every pair $(\xi_1, \xi_2)$ , there exists $\mu \in A$ such that $\mu \overline{\mu} = 1$ and $\beta \mu \xi_1 \overline{\xi}_2 + \beta \mu \xi_2 \overline{\xi}_1 = 0$ , which will make it possible to reduce to the case where $\beta = 0$ ; use prop. 5 of Chap. VI, § 2, No. 5.)

c) Show that if $u$ is normal, $U$ is the smallest convex set containing all the eigenvalues of $u$ . Give an example where $u$ is not normal but where $U$ still has the preceding property. (Take as eigenvalues of $u$ the elements $\pm 1 \pm i$ as simple eigenvalues, and 0 as a double eigenvalue.)

Remark 17

Suppose the conditions of No. 3 are satisfied; for every $\xi \in A$ , denote by $|\xi|$ the element $\rho \geqslant 0$ of K such that $\rho^2 = \xi \overline{\xi}$ (absolute value of $\xi$ ). For every square matrix $M = (\alpha_{ij})$ over A, put $f(M) = \max_i |\alpha_{ii}|$ , $g(M) = \max_{i,j} |\alpha_{ij}|$ , and denote by $\varphi(M)$ the largest absolute value of the eigenvalues of $M$ (one will show that this definition has a meaning when A is the field of quaternions over K).

a) Let $A, B, D$ be three square matrices of order $n$ over A. Show that if $D$ is diagonal, one has

g^2(ADB^*) \leq f^2(D)f(AA^*)f(BB^*)

(use inequality (1) of No. 1). Deduce that

g(ABB^*A^*) \leq f(AA^*)\varphi(BB^*)

(apply Prop. 5 to the hermitian matrix

BB^*

). Deduce that, for

m

arbitrary square matrices

A_i

(

1 \leq i \leq m

) over A, one has

g^2(A_1A_2\ldots A_m) \leq f(A_1A_1^*)\varphi(A_2A_2^*)\ldots\varphi(A_{m-1}A_{m-1}^*)f(A_m^*A_m)

\varphi^2(A_1A_2\ldots A_m) \leq \varphi(A_1A_1^*)\varphi(A_2A_2^*)\ldots\varphi(A_mA_m^*)

g^2(A_1A_2\ldots A_m) \leq \varphi(A_1A_1^*)\ldots\varphi(A_mA_m^*).

(For (3), reason by induction from (2). For (4), use Ex. 12; finally deduce (5) from (3).)

Show that inequality (3) no longer necessarily holds when $A_m^*A_m$ is replaced there by $A_mA_m^*$ (observe that one may have $f(A^*A) \neq f(AA^*)$ ), or when $\varphi(A_iA_i^*)$ is replaced by $f(A_iA_i^*)$ for $2 \leq i \leq m-1$ (take all the $A_i$ equal to the square matrix whose all elements are 1).

b) Let $u$ be a normal endomorphism of E; if $\lambda$ is an eigenvalue of $u$ (element of $K(i)$ when $A = K$ ), $\overline{\lambda}$ is an eigenvalue of $u^*u$ . For every normal matrix $M$ (matrix of a normal endomorphism of E with respect to an orthonormal basis), one therefore has $\varphi^2(M) = \varphi(MM^*)$ . Deduce that one also has $g(M) \leq \varphi(M)$ , and more generally, if $M_1, \ldots, M_m$ are normal,

g(M_1\ldots M_m) \leq \varphi(M_1)\ldots\varphi(M_m)

\varphi(M_1\ldots M_m) \leq \varphi(M_1)\ldots\varphi(M_m)

(use (4) and (5)).

c) Show that if $H$ is a positive hermitian matrix over A, one has $f(H) = g(H) \leq \varphi(H)$ (use (3) and b), by writing $H = AA^*$ .

Remark 18

The conditions of No. 3 are assumed to be satisfied.

a) For every endomorphism $u$ of $E$ , let $(e_i)$ be an orthonormal basis of $E$ formed by eigenvectors of $u^*u$ , and let $\rho_i$ be the eigenvalue of $u^*u$ corresponding to the vector $e_i$ ; put $s(u) = (\sum_{i=1}^n \rho_i)^{1/2}$ (square root of $\operatorname{Tr}(u^*u)$ when $A$ is commutative); one has $s(u^*) = s(u)$ . If $u, v$ are two endomorphisms of $E$ , $U, V$ their matrices with respect to the same orthonormal basis of $E$ , show that for the matrix $UV^* + VU^*$ , with elements in $K$ , one has $(\operatorname{Tr}(UV^* + VU^*))^2 \leq 4s(u)s(v)$ (observe that $(u^* + \lambda v^*)(u + \lambda v)$ is positive hermitian for every $\lambda \in K$ ); deduce that $s(u + v) \leq s(u) + s(v)$ . If in addition $A$ is commutative, one has

|\operatorname{Tr}(uv)|^2 \leq s(u)s(v) \quad \text{and} \quad |\operatorname{Tr}(u)| \leq \sqrt{n}s(u)

(write $u$ as a product by using Ex. 14 b)).

b) Suppose in addition that $A$ is commutative (thus equal to $K$ or $K(i)$ ). If $H_1, H_2$ are two square matrices hermitian positive, now prove that one has $\operatorname{Tr}(H_1H_2) \geq 0$ (reduce to the case where $H_2$ is a diagonal matrix). Deduce that one has (notations of Exercise 17)

|\operatorname{Tr}(H_1H_2)| \leq \varphi(H_1)\operatorname{Tr}(H_2) \leq \operatorname{Tr}(H_1)\operatorname{Tr}(H_2).

Conclude from these inequalities that for two arbitrary endomorphisms $u, v$ of $E$ , one has then $s(uv) \leq s(u)s(v)$ .

c) Assuming still that $A$ is commutative, let $\prod_{i=1}^n (x - \lambda_i)$ be the decomposition into linear factors of the characteristic polynomial of an arbitrary endomorphism $u$ of $E$ ; now prove that one has $\sum_{i=1}^n |\lambda_i|^2 \leq (s(u))^2$ and that, for the two members of this inequality to be equal, it is necessary and sufficient that $u$ be normal (use Exercise 12).

Remark 19

Suppose that the conditions of No. 3 are satisfied. Let $M$ be a square matrix of order $n$ over $A$ ; now prove that for every square submatrix $N$ of $M$ (obtained by deleting in $M$ a certain number of rows and the columns having the same indices as these rows), one has (notations of Exercise 17) $\varphi^2(N) \leq \varphi(MM^*)$ (apply suitably formula (4) of Exercise 17). If in particular $M$ is a normal matrix, $\varphi(N) \leq \varphi(M)$ (cf. Exercise 17 b)).

Remark 20

Suppose that the conditions of No. 3 are satisfied and in addition that $A$ is equal to $K$ or to $K(i)$ . Apply the results of Exercises 17 to 19 to the extension of $\Phi$ to the $p$ -th exterior powers (§ 1, No. 9) and to the $p$ -th exterior powers of the endomorphisms or matrices considered. In particular, now prove that, if $\prod_{i=1}^n (X - \lambda_i)$ and $\prod_{i=1}^n (X - \rho_i^2)$ are the decompositions into linear factors of the characteristic polynomials of an endomorphism $u$ of $E$ and of the endomorphism $u^*u$ , and if one supposes $|\lambda_i| \geqslant |\lambda_{i+1}|$ and $\rho_i \geqslant \rho_{i+1} \geqslant 0$ for $1 \leqslant i \leqslant n-1$ , one has

|\lambda_1 \lambda_2 \ldots \lambda_h| \leqslant \rho_1 \rho_2 \ldots \rho_h

for

1 \leqslant h \leqslant n-1

and

|\lambda_1 \lambda_2 \ldots \lambda_n| = \rho_1 \rho_2 \ldots \rho_n

Remark 21

Suppose that the conditions of No. 3 are satisfied. Let $u$ be a hermitian endomorphism of $E$ and let $\prod_{i=1}^n (X-\lambda_i)$ be the decomposition into linear factors of its characteristic polynomial; suppose $\lambda_i \geqslant \lambda_{i+1}$ for $1 \leqslant i \leqslant n-1$ .

a) Now prove that the greatest (resp. the least) of the eigenvalues $\lambda_i$ in $K$ is equal to the greatest (resp. the least) of the values of $\Phi(u(x), x)$ when $x$ ranges over the set of the $x \in E$ such that $\Phi(x, x) = 1$ . (Reason directly, or apply Exercise 16 c) by reducing to the case where $A = K(i)$ ( $§ 3$ , Exercise 4).)

b) Let $\Psi_v$ be the restriction to a vector subspace $V$ of $E$ of the hermitian form $\Psi'$ associated with $u$ , and let $u_v$ be the hermitian endomorphism of $V$ associated with $\Psi_v$ . Now prove that $\lambda_k$ is the smallest of the largest eigenvalues of the $u_v$ , when $V$ runs through the set of vector subspaces of $E$ of dimension $n-k+1$ (use prop. 5).

Remark 22

Suppose that the conditions of No. 3 are fulfilled, and in addition that $A$ is equal to $K(i)$ or to the field of quaternions over $K$ . Let $u, v$ be two normal endomorphisms of $E$ ; let $(E_i)_{1 \leqslant i \leqslant r}$ (resp. $(F_j)_{1 \leqslant j \leqslant s}$ ) be the decomposition of $E$ into a direct sum of pairwise orthogonal subspaces, such that in $E_i$ (resp. $F_j$ ) there is an orthonormal basis formed of eigenvectors of $u$ (resp. $v$ ) for the same eigenvalue $\lambda_i$ (resp. $\mu_j$ ), and that $\lambda_h$ and $\lambda_i$ (resp. $\mu_j$ and $\mu_k$ ) are not transformed into one another by an inner automorphism of $A$ if $h \neq i$ (resp. $j \neq k$ ) (cf. prop. 4 and exerc. 9). In order that an endomorphism $w$ of $E$ be such that $uw = vw$ , it is necessary and sufficient that for every $j$ ( $1 \leqslant j \leqslant s$ ), $w(F_j)$ be contained in one of the $E_i$ and that the image by $w$ of every eigenvector of $v$ corresponding to the eigenvalue $\mu_j$ be an eigenvector of $u$ corresponding to the eigenvalue $\mu_j$ (which in particular implies that $\mu_j$ and $\lambda_i$ are transformed into one another by an inner automorphism of $A$ ). Deduce that if this is so, then $u^*w = wv^*$ .

Remark 23

Suppose that the conditions of No. 3 are fulfilled. Let $u$ and $v$ be two endomorphisms of $E$ .

a) Suppose that $u$ and $uv$ are normal. In order that $vu$ be normal, it is necessary and sufficient that $v$ and $u^*u$ be commutable. (To see that the condition is necessary, use the relation $u(vu) = (uv)u$ and exerc. 22; to see that it is sufficient, use exerc. 14 a) and 10 d).)

b) Suppose that $u, v$ and $uv$ are normal; let $\beta$ be the greatest eigenvalue of $uu^*$ , and let $F$ be the subspace of $E$ formed by the eigenvectors of $uu^*$ corresponding to the eigenvalue $\beta$ . Now prove that $v(F) \subset F$ . (Remark that for every normal endomorphism $u$ and every $x \in E$ , one has

\Phi(u(x), u(x)) = \Phi(u^*(x), u^*(x)).

)

Reduce to the case where A is equal to K(i) or to the quaternion field over K; F then admits a basis formed by eigenvectors for u (and u*); remark that for such a vector z, one has $\Phi(u^*u v(z), v(z)) = \beta \Phi(v(z), v(z))$ . Deduce that νu is normal (reason by induction on the number of distinct eigenvalues of $u^*u$ ). If h (resp. $h'$ ) is the positive hermitian endomorphism such that $h^2 = uu^*$ (resp. ${h'}^2 = \nu \nu^*$ ) and if one sets $u = hu_1, \nu = h'\nu_1$ , where $u_1$ and $\nu_1$ are unitary, h permutable with $u_1$ and $h'$ with $\nu_1$ (exerc. 14 a)), prove that the pairs $(h, h'), (h, \nu_1)$ and $(h', u_1)$ are permutable; converse. Deduce that $u^m \nu^n, \nu^n u^m, u \nu^*$ and $\nu^* u$ are then normal ( $m$ and $n$ arbitrary integers $> 0$ ).

Remark 24

Suppose that the conditions of No. 3 are fulfilled. Let $\Gamma$ be a group of automorphisms of E such that every $u \in \Gamma$ is normal. Now prove that there exists a decomposition of E into direct sum of subspaces $E_k$ ( $1 \leq k \leq r$ ) pairwise orthogonal, such that the restriction to $E_k$ of every $u \in \Gamma$ is of the form $\lambda_k \nu_k$ , where $\lambda_k$ is an element $> 0$ of K and where $\nu_k$ is a unitary endomorphism of $E_k$ (decompose each $u \in \Gamma$ in the form $h \nu$ , where $\nu$ is unitary, h positive hermitian and $h^2 = uu^*$ (exerc. 14 a)); use exerc. 23 b) and apply cor. 2 of th. 2 to the algebra generated by the hermitian endomorphisms h corresponding to the $u \in \Gamma$ . Deduce that if $\Gamma$ is finite, the elements $u \in \Gamma$ are unitary.

Remark 25

Suppose that $A = K$ (maximal ordered field). Let L be an euclidean space of dimension $n$ over A, whose metric form is positive non-degenerate.

Let S be a non-degenerate affine quadric in L (§ 6, exerc. 25). If S admits a center $a$ and if one takes $a$ for origin in L, now prove that there exists an orthonormal basis $(e_i)$ for L such that, with respect to this basis, S has for equation $\lambda_1 \xi_1^2 + \cdots + \lambda_n \xi_n^2 = 1$ . Moreover, if two orthonormal bases have this property, the elements $\lambda_i \in K$ corresponding to them are the same up to order.

If S has no center, show that there exists a point $b$ of S and (taking $b$ as origin) an orthonormal basis for L such that, with respect to this basis, S has the equation $\lambda_1 \xi_1^2 + \cdots + \lambda_{n-1} \xi_{n-1}^2 + \xi_n = 0$ . (If $\overline{S}$ is the projective quadric such that $S = L \cap \overline{S}$ , determine $b$ by the condition that the line passing through $b$ and $c$ be perpendicular to the tangent hyperplane to S at the point $b$ , where c is the (at infinity) pole with respect to $\overline{S}$ of the hyperplane at infinity $H_0$ ).

Remark 26

a) Let K be a commutative field, S a subset of K such that K is a maximal S-orderable field (chap. VI, § 2, exerc. 8). Let $f$ be a polynomial of $K[X]$ , L its field of roots. Show that if, for every (total) order structure on K compatible with its ring structure, L admits a structure of an ordered extension of K, then necessarily $L = K$ . (Argue by contradiction: proceeding as in exerc. 8 e) of chap. VI, § 2, reduce to the case where one would have $[L : K] = 2$ . Conclude by observing that if $b \in K$ is not a square, there exists a total order structure on K, compatible with its ring structure, for which $b < 0$ (cf. chap. VI, § 2, no. 3, lemma of th. 1)).

b) Extend the results of No. 3 (with the exception of prop. 6) to the case where K is a maximal S-orderable field (). (Begin by proving the analogue of prop. 5, using a). Establish next the analogue of cor. 2 of th. 2 for the case where the commutative algebra B consists of hermitian endomorphisms, using prop. 10 of chap. VIII, § 9, no. 4. Pass to the case of a normal endomorphism u for A = K(i), observing that one can then write $u = v + i w$ , where v and w are hermitian and permutable. Finally, if A is the field of quaternions over K, E0 the space E considered as a vector space of dimension 2n over K(i), and if one sets $\Phi(x, y) = \Phi_1(x, y) + \Phi_2(x, y)i$ , where $\Phi_1$ and $\Phi_2$ take their values in K(i), observe that if u is normal for $\Phi$ , it is also normal for $\Phi_1$ .)

Remark 27

Let K be a maximal ordered field, E a vector space of dimension n over K, $\Phi$ a non-degenerate symmetric bilinear form on E, of signature (s, t) distinct from (n, 0) and from (0, n). Let $(e_i)$ be an orthogonal basis for $\Phi$ , such that $\Phi(e_i, e_i) = 1$ for $1 \leq i \leq s$ , $\Phi(e_i, e_i) = -1$ for $s+1 \leq i \leq n$ . For every orthogonal transform $u \in \mathbf{U}(\Phi)$ , let

U = \begin{pmatrix} M & N \\ P & Q \end{pmatrix}

be the matrix of u with respect to $(e_i)$ , written in the form of a square tableau of matrices corresponding to the partition of $(1, n)$ into $(1, s)$ and $(s+1, n)$ .

a) Prove the relations

\begin{align*} {}^tM \cdot M - {}^tP \cdot P &= I_s \\ {}^tQ \cdot Q - {}^tN \cdot N &= I_t \\ {}^tM \cdot N - {}^tP \cdot Q &= 0. \end{align*}

b) Let R be a matrix over K with t rows and s columns, such that $I_s - {}^tR \cdot R$ is the matrix of a positive non-degenerate hermitian form (on $K^s$ ). Show that $\det(M + \lambda NR)$ does not change sign for $-1 \leq \lambda \leq 1$ in K (show, using a), that ${}^t(M + \lambda NR)(M + \lambda NR)$ is the matrix of a positive non-degenerate symmetric form).

c) Put $\sigma(u) = \sigma(U) = \operatorname{sgn}(\det M)$ ; show that, for any two elements $u, v$ of the orthogonal group $\mathbf{U}(\Phi)$ , one has $\sigma(uv) = \sigma(u)\sigma(v)$ (use a) and b)). Deduce that the commutator group of the special orthogonal group $\mathbf{SU}(\Phi)$ is distinct from $\mathbf{SU}(\Phi)$ (cf. § 10, exerc. 9).