Let \(f\) be a symmetric bilinear form on \(\RR^n\). Use the result of Exercise 8 to prove the existence of an ordered basis \(\mathfrak{B}\) such that \([f]_{\mathfrak{B}}\) is diagonal.
Solution
Let \[ q(v)=f(v,v) \] be the quadratic form associated with the symmetric bilinear form \(f\).
1. Apply Exercise 8 to \(q\).
Exercise 8 provides an invertible linear operator \[ U:\mathbb R^n\longrightarrow\mathbb R^n \] and scalars \[ c_1,\ldots,c_n\in\{-1,0,1\} \] such that \[ (U^\dagger q)(x_1,\ldots,x_n) =q(Ux) =\sum_{i=1}^n c_i x_i^2. \]
2. Use the columns of \(U\) as the desired basis.
Let \(e_1,\ldots,e_n\) be the standard basis and set \[ u_i=Ue_i. \] Since \(U\) is invertible, \[ \mathfrak B=(u_1,\ldots,u_n) \] is an ordered basis.
The symmetric bilinear form associated with the pulled-back quadratic form \(U^\dagger q\) is \[ h(x,y)=f(Ux,Uy). \] By polarization of \[ U^\dagger q(x)=\sum_i c_i x_i^2, \] we obtain \[ h(x,y)=\sum_{i=1}^n c_i x_i y_i. \] Hence \[ h(e_i,e_j)=c_i\delta_{ij}. \] But \[ h(e_i,e_j)=f(Ue_i,Ue_j)=f(u_i,u_j). \] Therefore \[ \boxed{ [f]_{\mathfrak B} =\operatorname{diag}(c_1,\ldots,c_n),} \] which is diagonal.