I'm confused about this.
Consider a bilinear symmetric form $\phi :V\times V\rightarrow \mathbb{R}$ and its associated quadratic form $Q$. Then considering a generic vector subspace $W\subseteq V$, what is exactly inside the vector subspace $W \cap W^{\bot}$? Of course if $\phi$ is positive definite $W \cap W^{\bot}= \big\{ \vec{o} \big\}$.
But if $\phi$ is semipositive or indefinite can I say that $W \cap W^{\bot}= \big\{ \vec{x} \in W | \phi(\vec{x},\vec{x})=0 \big\}$?
(In this case $W \cap W^{\bot}$ is a subspace of the set of the isotropic vectors $\mathscr{I}$ )
But are there only those vectors? Is it better to write that
$$W \cap W^{\bot}= \big\{ \vec{x} \in W | \phi(\vec{x},\vec{y})=0, \forall \vec{y} \in W\big\}?$$
And what are the difference between these two sets? Which is the right one?
What I think is that I can only say that $W \cap W^{\bot} \supseteq \big\{ \vec{x} \in W | \phi(\vec{x},\vec{x})=0 \big\}$ because there could be other vectors with respect to wich $\vec{x}$ must be orthogonal, besides himself for $\vec{x}$ to be in $W \cap W^{\bot}$ but I can't really understand if this is right.
By the Cauchy-Schwarz inequality (which also holds for semidefinite bilinear forms), we have $$|\phi(x,y)|^2 \le \phi(x,x) \, \phi(y,y).$$ Hence, $$\{x \in W : \phi(x,x) = 0\} = \{x \in W : \phi(x,y) = 0 \; \forall y \in W\}.$$