We fix two symplectic forms $\omega$ and $\omega _ { 1 }$ on $E$, and let $u \in \mathrm{GL}(E)$ be the unique automorphism such that $\omega_1(x,y) = \omega(u(x),y)$ for all $(x,y) \in E^2$. We assume that $E$ is of dimension 4 and that $u$ has no real eigenvalue. Let $\mathcal{B}$ be a basis of $E$ such that $\operatorname{Mat}_{\mathcal{B}}(\omega) = J_4$, and let $Z, Y \in \mathbb{C}^4$ be eigenvectors as in question 15, with $z_1, z_2, y_1, y_2$ as defined in question 16. Show that, by replacing $Y$ with $\xi Y$ where $\xi \in \mathbb { C } \backslash \{ 0 \}$ is suitably chosen, we have $\omega \left( z _ { 1 } , y _ { 1 } \right) = - 1$ and $\omega \left( z _ { 1 } , y _ { 2 } \right) = 0$.
We fix two symplectic forms $\omega$ and $\omega _ { 1 }$ on $E$, and let $u \in \mathrm{GL}(E)$ be the unique automorphism such that $\omega_1(x,y) = \omega(u(x),y)$ for all $(x,y) \in E^2$. We assume that $E$ is of dimension 4 and that $u$ has no real eigenvalue. Let $\mathcal{B}$ be a basis of $E$ such that $\operatorname{Mat}_{\mathcal{B}}(\omega) = J_4$, and let $Z, Y \in \mathbb{C}^4$ be eigenvectors as in question 15, with $z_1, z_2, y_1, y_2$ as defined in question 16.
Show that, by replacing $Y$ with $\xi Y$ where $\xi \in \mathbb { C } \backslash \{ 0 \}$ is suitably chosen, we have $\omega \left( z _ { 1 } , y _ { 1 } \right) = - 1$ and $\omega \left( z _ { 1 } , y _ { 2 } \right) = 0$.