Theorem 7

Let E be a vector space over a field K, F a subspace of E and F' the orthogonal of F in E.*

(i) $\dim F' \geq \operatorname{codim}_E F;$ for $\dim F'$ to be finite, it is necessary and sufficient that $\operatorname{codim}_E F$ be finite, and then $\dim F' = \operatorname{codim}_E F$.

(ii) The orthogonal of F' in E is equal to F.

(iii) Every finite-dimensional subspace G' of E is the orthogonal of some subspace of E, necessarily equal to the orthogonal of G' in E and of finite codimension.*

(i) We know that F' is isomorphic to the dual (E/F)* (§ 2, no. 6, Corollary to Proposition 9) and hence the assertion follows from Theorem 4, since $\dim(E/F) = \operatorname{codim}_E F$ by definition.

(ii) Let $F_1$ be the orthogonal of F' in E; clearly $F \subset F_1$ and the orthogonal $F'_1$ of $F_1$ is equal to F' (§ 2, no. 4); the canonical linear mapping $(E/F_1)^* \to (E/F)^*$, the transpose of $E/F \to E/F_1$ is therefore bijective (§ 2, no. 6, Corollary to Proposition 10); it then follows from Proposition 10 that the canonical mapping $E/F \to E/F_1$ is bijective, which implies $F_1 = F$.

(iii) Let G' be a subspace of E* of finite dimension $p$ and let F be its orthogonal in E; then $\operatorname{codim}_E F \leq \dim G'$. For, if $(a_i^*)_{1 \leq i \leq p}$ is a basis of G', F is the kernel of the linear mapping $x \mapsto (\langle x, a_i^* \rangle)$ from E to $K_s^p$ whose rank is at most $p$ (no. 4, Proposition 9), whence the conclusion (no. 4). Then let F' be the orthogonal of F in E*; it follows from (i) that $\dim F' \leq \dim G'$; but on the other hand obviously $G' \subset F'$, whence $F' = G'$ (§ 2, no. 3, Corollary 4 to Proposition 4).