Let \(A\) be a symmetric \(n \times n\) matrix over \(\RR\), and let \(q\) be the quadratic form on \(\RR^n\) given by \[ q(x_1, \ldots, x_n) = \sum_{i,j} A_{ij}x_ix_j. \] Generalize the method used in Exercise 7 to show that there is an invertible linear operator \(U\) on \(\RR^n\) such that \[ (U^{\dagger}q)(x_1, \ldots, x_n) = \sum_{i=1}^{n} c_ix_i^2 \] where \(c_i\) is \(1\), \(-1\), or \(0\), \(i = 1, \ldots, n\).
Solution
Let \(f\) be the symmetric bilinear form associated with \(q\), so \[ q(v)=f(v,v). \] We prove by induction on \(n=\dim\mathbb R^n\) that there is a basis \(u_1,\ldots,u_n\) satisfying \[ f(u_i,u_j)=0\quad(i\ne j), \qquad f(u_i,u_i)\in\{-1,0,1\}. \]
1. The assertion holds in dimension zero, and also when \(q\) is identically zero.
If \(n=0\) there is nothing to prove. For arbitrary \(n\), if \(q=0\), then polarization gives \[ f(v,w)=\frac12\bigl(q(v+w)-q(v)-q(w)\bigr)=0 \] for all \(v,w\). Thus every basis has Gram matrix zero, and we may take \(c_1=\cdots=c_n=0\).
2. If \(q\) is not zero, one can split off a vector of value \(1\) or \(-1\).
Choose \(v\in V\) with \[ q(v)=f(v,v)\ne0. \] Let \[ c=\operatorname{sgn}(q(v))\in\{-1,1\}, \qquad u_1=\frac{v}{\sqrt{|q(v)|}}. \] Then \[ f(u_1,u_1)=c. \]
Define \[ W=\{w\in V:f(u_1,w)=0\}. \] For any \(x\in V\), put \[ w=x-c\,f(u_1,x)u_1. \] Since \(c^2=1\), \[ f(u_1,w) =f(u_1,x)-c f(u_1,x)f(u_1,u_1) =0. \] Hence \[ x=c f(u_1,x)u_1+w \] with \(w\in W\). Also \[ \mathbb R u_1\cap W=0 \] because \(f(u_1,u_1)=c\ne0\). Therefore \[ V=\mathbb R u_1\oplus W, \qquad \dim W=n-1. \]
3. Apply induction to the orthogonal complement.
The restriction of \(f\) to \(W\) is again a symmetric bilinear form, with quadratic form \(q|_W\). By the induction hypothesis, \(W\) has a basis \[ u_2,\ldots,u_n \] such that \[ f(u_i,u_j)=0\quad(2\le i\ne j\le n) \] and \[ f(u_i,u_i)=c_i\in\{-1,0,1\} \qquad(2\le i\le n). \] Since every \(u_i\) with \(i\ge2\) lies in \(W\), \[ f(u_1,u_i)=0. \] Thus \(u_1,\ldots,u_n\) is a basis of \(V\) with diagonal Gram matrix \[ \operatorname{diag}(c_1,\ldots,c_n), \qquad c_i\in\{-1,0,1\}, \] where \(c_1=c\).
4. Convert the constructed basis into the required operator \(U\).
Let \(e_1,\ldots,e_n\) be the standard basis and define the invertible linear operator \(U\) by \[ Ue_i=u_i. \] For \[ x=\sum_{i=1}^n x_i e_i, \] we have \[ Ux=\sum_i x_i u_i. \] Using orthogonality of the \(u_i\), \[ \begin{aligned} (U^\dagger q)(x) &=q(Ux)\\ &=f\left(\sum_i x_i u_i,\sum_j x_j u_j\right)\\ &=\sum_{i=1}^n c_i x_i^2. \end{aligned} \] Therefore \[ \boxed{ (U^\dagger q)(x_1,\ldots,x_n) =\sum_{i=1}^n c_i x_i^2, \qquad c_i\in\{-1,0,1\}.} \]