Let \(q\) be the quadratic form on \(\RR^2\) given by \[ q(x_1, x_2) = 2bx_1x_2. \] Find an invertible linear operator \(U\) on \(\RR^2\) such that \[ (U^{\dagger}q)(x_1, x_2) = 2bx_1^2 - 2bx_2^2. \]
Solution
Define \[ U(x_1,x_2)=(x_1+x_2,\ x_1-x_2). \]
1. The operator \(U\) is invertible.
Its matrix in the standard basis is \[ [U]= \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}, \] whose determinant is \[ -2\ne0. \] Therefore \(U\) is invertible.
2. Compute the pulled-back quadratic form.
By definition, \[ (U^\dagger q)(x_1,x_2) =q(U(x_1,x_2)). \] Since \[ q(z_1,z_2)=2bz_1z_2, \] we get \[ \begin{aligned} (U^\dagger q)(x_1,x_2) &=2b(x_1+x_2)(x_1-x_2)\\ &=2b(x_1^2-x_2^2)\\ &=2bx_1^2-2bx_2^2. \end{aligned} \] Thus \[ \boxed{(U^\dagger q)(x_1,x_2)=2bx_1^2-2bx_2^2.} \]