$(V, \omega)$ symplectic vector space a$W \subseteq V$ linear subspace. We define the reduced space $(\bar W , \bar \omega)$ as $\bar W = W/(W \cap W^\omega)$, equipped with the form $\bar\omega ([v], [w]) := \omega(v, w)$ for all $v, w \in W$. Prove that $\bar\omega$ is a symplectic form
I have already shown that $\bar \omega$ is well-defined, bilinear,skew-symmetric, but I am unsure on how to prove $\bar \omega$ is non degenerate
My try:
Let $[w]\in \ker \bar \omega \iff \bar\omega([w],[v])=0, \forall [v]\in \bar W$ $\iff \omega(w,v)=0, \forall v \in W$ ...(*)
$\iff w \in \ker \omega $
I am unsure about (*), can I go from $\forall [v]\in \bar W$ to $\forall v\in W$? and how do I proceed from here?
For reference:
Here are more steps: fix $[v] \in \bar{W}$, and assume that $\bar{\omega}([v],\cdot) = 0$. We will show that $\omega(v,w) = 0$ for all $w\in W$. To do so, we write $\omega(v,w) = \bar{\omega}([v],[w]) = 0$, by assumption. This means that $v\in W^\omega$. But we had $v\in W$ to begin with (or else $[v]$ would not make sense), so $v\in W\cap W^\omega$ and therefore $[v] = [0]$ by definition of quotient (we have shown that $\ker\bar{\omega} = 0$, i.e., $\bar{\omega}$ is non-degenerate).