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

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. $u^* = -u$).