Presymplectic vector space and symplectic subspaces

Question

Let $(V,\omega)$ be a finite dimensional presymplectic vector space and $W, U$ be symplectic subspaces. Is then $span(W,U)=\{z\in V|\exists u\in U, w \in W: z=w+u \}$ also a symplectic subspace ?

Answer 1

Consider $V=\mathbb{R}^4$ with the symplectic form $\omega$ defined by the following matrix with respect to the Cartesian basis $e_1$, $e_2$, $e_3$, $e_4$:
$$ \omega=\begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}. $$ Let $W:= span\{e_1, e_2\}$ and $U:= span\{e_2,e_3\}$. Then both $W$ and $U$ are symplectict but $U+W=span\{e_1, e_2, e_3\}$ is clearly not symplectic.

EDIT (reaction to a comment): Consider $V'=\mathbb{R}^6$ with the symplectic form $\omega'$ defined by the following matrix with respect to the Cartesian basis $e_1$, $e_2$, $e_3$, $e_4$, $e_5$, $e_6$: $$\omega' = \begin{pmatrix} 0 & 1 & 0 & -1 & 0 & 1 \\ -1 & 0 & 1 & 0 & 1 & 0 \\ 0 & -1 & 0 & 1 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$ Let $W':=span\{e_1,e_2\}$ and $U'=span\{e_3,e_4\}$. Then both $W'$ and $U'$ are symplectic subspaces of $V'$, it holds $W'\cap U' = 0$ because $e_1$, $e_2$, $e_3$, $e_4$ are linearly independent, but $W'+U'\subset V'$ is not a symplectic subspace.