Problem: Consider a real vector space $V$ of dimension $2m,\ m\in\mathbf{Z}^+$ and a non-singular quadratic form $Q$ on $V$. Suppose $Q$ vanishes on a subspace of dimension $m$, what is the signature of $Q$?
Attempts made: by Sylvester's Law of Inertia, choose a basis {$a_1,..a_m,b_1,..b_m$} such that the quadratic form is diagonal (up to reordering) with $p$ 1's, $q$ -1's, and n-(p+q) zeros on the diagonal. If it vanishes on subspace $U$ of dimension $m$, we let {$b_1,..b_m$} be a basis for $U$. We can characterise $p,q$ as the dimensions of the largest subspaces on which $Q$ is positive (negative) definite, but I am unsure how to proceed.
Over any field, whenever $\dim(V)=2m$ and a quadratic form vanishes on a subspace of dimension $m$, the form is hyperbolic (and the subspace is called a Lagrangian). In that case, over an ordered field, the signature of the quadratic form is $0$.
A possible way to see this is to take a basis of a Lagrangian $L$, say $(e_1,\dots,e_m)$, and then choose some $f_i\in V$ which is not orthogonal to $e_i$, and is orthogonal to the $f_j$ for $j<i$ (try to think about why that exists).
Then if we define $W_i=\langle e_i, f_i\rangle$, we get an orthognal sum $V=W_1\perp\dots\perp W_m$, and each $W_i$ is an isotropic plane. Now try to see that an isotropic plane (which is necessarily hyperbolic) has signature $0$.