Totally isotropic subspace

A subspace $E$ is totally isotropic for $B$ when $B(u,v)=0$ for all $u,v\in E$. For a real nondegenerate bilinear form of inertia $(p,q)$, its dimension is at most $\min (p,q)$.

For a nondegenerate real quadratic form of inertia $(p,q)$, projection of a totally isotropic subspace onto either its positive or negative coordinate subspace is injective: a vector with one projection zero would have strictly one-signed quadratic value unless it vanished. Thus its dimension is at most both $p$ and $q$. This bound is attained by pairing one positive and one negative coordinate for each of $\min (p,q)$ independent null vectors.

A quadratic form of rank $r$ on ${\mathbb{C}}^n$ is congruent to $x_1^2+\cdots +x_r^2$. Pair its nonzero coordinates into vectors $e_{2j-1}+i{e}_{2j}$, and include the radical of a bilinear form. Their span is a totally isotropic subspace and has dimension $\lfloor r/2\rfloor +n-r=n-\lceil r/2\rceil \ge \lfloor n/2\rfloor$. The form vanishes on every vector of this subspace, not merely on its basis vectors.

Any $d$ quadratic forms on ${\mathbb{C}}^n$ have a common nonzero zero when $n\ge 2^d$. Choose a totally isotropic subspace for the first form, of dimension at least $\lfloor n/2\rfloor$. Restrict the remaining forms to it and repeat by induction. After $d$ stages at least one dimension remains. This elementary sufficient bound does not assert that $2^d$ is the best possible threshold.

For a real quadratic form on an $n$-dimensional vector space, with rank $r$ and signature $s$ defined as positive minus negative index, the maximum dimension of a totally isotropic subspace is $n-(r+|s|)/2$. Quotienting by the radical of a bilinear form leaves a nondegenerate form of indices $p,q$. Projection of an isotropic subspace to each sign-coordinate space is injective, bounding its dimension by $\min (p,q)$. The radical and vectors pairing one positive with one negative unit coordinate attain the bound $n-r+\min (p,q)$.