Bourbaki

Theorem 2

Tag 0DM5, permanent. Cite it as [Bourbaki, Tag 0DM5].

Algebra, chapter IX, § 7. Formes hermitiennes et corps ordonnés

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 Φ(y,u(x))=Φ(u∗(y),x)=0\Phi(y, u(x)) = \Phi(u*(y), x) = 0 (resp. Φ(y,u(x))=Φ(u∗(y),x)‾=0\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 Φ\Phi) is identical with the restriction to V⁰ of the adjoint of u.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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