Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 10E / Solution

A quadratic form over $\mathbb{R}$ or $\mathbb{C}$ is a function $q(x)=B(x,x)$ for a symmetric bilinear form $B$, equivalently a homogeneous polynomial of degree two. Its polarization is $B(x,y)=[q(x+y)-q(x)-q(y)]/2$. If $q$ is not identically zero, choose $v$ with $q(v)\ne 0$. Every $x$ decomposes aswhere the second term lies in $v^{\perp }$ and $kv\cap v^{\perp }=\{0\}$. The quadratic form is consequently a direct sum of its one-dimensional value on $v$ and its restriction to $v^{\perp }$. Induction on dimension diagonalizes it. If $q=0$, any basis suffices. Rescale every nonzero diagonal coefficient: over the reals it becomes $1$ or $-1$, while over the complexes a square root makes it $1$. HenceOver $\mathbb{C}$ the negative coefficients can all be avoided.

For a real quadratic form, let $p$ and $m$ be its positive and negative indices in a diagonalization. Its rank of a quadratic form is $p+m$ and its signature of a quadratic form is $p-m$, or the ordered pair $(p,m)$ in the alternative convention. These indices are intrinsic: the largest dimension of a subspace where the form is positive definite is $p$, since any larger subspace intersects the nonpositive coordinate subspace nontrivially; similarly for $m$. For the form here, completing squares givesThe two displayed linear coordinates, together with $x_1$, form an invertible coordinate change. ThereforeThere is one zero coordinate direction.

The PDF's final dimension condition is $n\ge 2^d$, not the fraction produced by the TeX conversion. We prove the requested common zero of complex quadratic forms by induction on $d$. A complex quadratic form of rank $r$ has canonical expression $x_1^2+\cdots +x_r^2$. Pair its nonzero coordinates into vectors $e_{2j-1}+i{e}_{2j}$ and add the $n-r$ radical coordinate vectors. Their span is a totally isotropic subspace, because the form is zero on every linear combination. This proves that complex quadratic forms have a large isotropic subspace: its dimension isFor $d=0$ there is a nonzero vector whenever $n\ge 1$. For $d\ge 1$, take such a subspace $E$ for $q_1$. If $n\ge 2^d$, then $\dim E\ge 2^{d-1}$. Apply the induction hypothesis to the restrictions of $q_2,\dots ,q_d$ to $E$. It supplies a nonzero $x\in E$ annihilated by those forms, and $q_1(x)=0$ by construction. ThusThe argument makes no nondegeneracy assumption on any of the forms.