Consider a symplectic vector space $(\mathbb{R}^{2n}, \omega_0)$ with standard symplectic form $\omega_0$ defined by: $$\omega_0(x,y) = xJ_0y^T$$ where $J_0=\begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}$. Let $W_1, W_2$ be symplectic subspaces of $\mathbb{R}^{2n}$ such that $W_1\cap W_2 = \{0\}$. Is $W_1 + W_2$ also symplectic?
It seems to be true but neither can I give a proof nor a counter example.
I got a counter example. Take $$W_1= span\{(1,0,0,0,0,0), (0,0,0,1,0,0)\}$$ and $$W_2= span\{(1,1,0,0,0,0), (0,0,0,1,0,1)\}$$. $W_1,W_2$ are symplectic because if $e_1, e_2$ denotes the basis elements of either $W_1$ or $W_2$, then $ \omega_0(e_1,e_2)= 1\neq 0$. i.e., $\omega_0$ is non-degenerate on both $W_1$ and $W_2$. Clearly $W_1 \cap W_2 = \{0\}.$ But $$W_1 + W_2 = span \{(1,0,0,0,0,0), (0,0,0,1,0,0),(1,1,0,0,0,0), (0,0,0,1,0,1)\}$$ is not symplectic since $(0,1,0,0,0,1) \in W_1 + W_2$ and $\omega_0((0,1,0,0,0,1), x)=0$ $\forall x\in W_1 +W_2$.(i.e., $\omega_0$ is degenerate on $W_1 + W_2$.)
Furthermore, this will be true if $W_1 \subset W_2^{\omega_0}$ and $W_2 \subset W_1^{\omega_0}$. These conditions are sufficient but not neccessary.