Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 2 / 12E (Linear Algebra)

Associate to $Q$ its symmetric bilinear form by the polarization identity,We first construct a diagonal basis by induction on the dimension. A zero quadratic form is already diagonal. Otherwise choose $v$ with $Q(v)\ne 0$. Every $x$ has the decompositionwhose second term lies in $W=\{w:B(w,v)=0\}$. Also $\mathbb{R}v\cap W=\{0\}$, so $V=\mathbb{R}v\oplus W$ and $\dim W=n-1$. Apply the induction hypothesis to the restricted quadratic form on $W$. The cross terms with $v$ vanish, giving a diagonal basis on $V$. Rescale each vector with diagonal coefficient $a\ne 0$ by $1/\sqrt{|a|}$, and reorder positive, negative and zero coefficients. This yieldsNo nondegeneracy assumption was used.

For uniqueness, let $V_{+}$ be the positive coordinate subspace and $V_{\le 0}$ the span of all negative and zero coordinate vectors. Their dimensions are $p$ and $n-p$. Any subspace $W$ on which $Q$ has positive definiteness intersects $V_{\le 0}$ only in zero. The dimension formula for a sum of subspaces therefore gives $\dim W\le p$. Conversely $V_{+}$ itself has dimension $p$ and positive restriction. Thus $p$ is the largest possible dimension of a positive subspace, an intrinsic description independent of the basis. Applying the same argument to $-Q$ shows that $q$ is the largest dimension of a negative subspace. This proves Sylvester's law of inertia, including uniqueness of the zero count $n-p-q$.

The extremal subspaces themselves need not be unique. For $Q(x,y)=x^2-y^2$, the coordinate axes give one positive and one negative line. For any $t\ne 0$, putThen $Q(u)=1$, $Q(v)=-1$, $B(u,v)=0$, and the determinant of the change of basis is one. The distinct lines $\mathbb{R}u$ and $\mathbb{R}v$ give another positive/negative decomposition, with the same intrinsic counts $p=q=1$.