Remark 3

It is easily verified that, for $x, y$ in $E$ and $u \in \Lambda(\Phi)$, one has $\Phi(u(x), y) = \Phi(x, \overline{u}(y))$. Thus the adjoint endomorphism of a direct similitude $u$ is the direct similitude $\overline{u}$ conjugate of $u$ in $\Lambda(\Phi)$.