Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4 / 10G / Solution

Two real symmetric bilinear forms are congruent if an invertible linear map takes one to the other; in matrices, $B=S^{\mathsf{T}}AS$ for some invertible $S$. The spectral theorem for real symmetric matrices gives an orthogonal eigenbasis. Rescaling each basis vector with nonzero eigenvalue then reduces any such form toSylvester's law of inertia asserts that $(p,q,r)$ is independent of the congruence reduction. To prove it, $p$ is the greatest dimension of a subspace on which the form is positive definite: the positive coordinate subspace attains it, whereas any subspace of dimension greater than $p$ meets the nonpositive coordinate subspace of dimension $n-p$ and therefore cannot be positive definite. Applying the same argument to the negative of the form determines $q$. Then $r=n-p-q$ is determined too.

The rank of a quadratic form is $p+q$, and its signature of a quadratic form is $p-q$ under the signed-difference convention. They determine $p,q,r$ and hence the congruence class. Counting all pairs $p,q\ge 0$ with $p+q\le n$ givesFor $n\ge 1$, exactly two classes have every level set bounded: the positive-definite and negative-definite classes. Their nonempty nonzero levels are ellipsoids, and their zero level is the origin. A degenerate form has an entire null line in its zero level. An indefinite nondegenerate form also has a nonzero isotropic vector, for example the sum of one positive and one negative normalized basis vector, whose scalar multiples make the zero level unbounded.

For the displayed three-dimensional form,Its leading principal minors are $3,14,32$, all positive. Sylvester's criterion makes it positive definite, so the surface is a bounded ellipsoid. Directly, its quadratic expression iswhich also proves positivity and boundedness of the unit level.