Remark 9

Let $A$ be a field, $E_1, E_2$ two vector spaces not reduced to 0 over $A$, $\Phi_1$ (resp. $\Phi_2$) a non-degenerate sesquilinear form on $E_1$ (resp. $E_2$) for an antiautomorphism $J_1$ (resp. $J_2$) of $A$. Let $u$ be a linear mapping of $E_1$ onto $E_2$ such that the relation $\Phi_1(x, y) = 0$ implies $\Phi_2(u(x), u(y)) = 0$.

a) Show that $u$ is a bijection of $E_1$ onto $E_2$. (If $u(0)$ were not reduced to 0, show that there would exist in $E_1$ two vectors $a, b$ such that $u(a) \neq 0, u(b) = 0$ and $\Phi_1(a, b) \neq 0$; if $H$ is the hyperplane of the $x \in E_1$ such that $\Phi_1(a, x) = 0$, remark that one would have $u(H) = E_2$.

b) Show that if dim $E_1 \geqslant 2$, there exists $\alpha \in A$ such that one has $\Phi_2(u(x), u(y)) = \Phi_1(x, y)\alpha$ whatever $x, y$ in $E_1$. (For every $y \in E_1$, show that there exists an element $m(y) \in A$ such that $\Phi_2(u(x), u(y)) = \Phi_1(x, y)m(y)$ for every $x \in E_1$, and that if $y$ and $y'$ are linearly independent in $E_1$, one has $m(y + y') = m(y) = m(y')$).