Let \(q\) be the quadratic form on \(\RR^2\) given by \[ q(x_1, x_2) = ax_1^2 + 2bx_1x_2 + cx_2^2, \quad a \neq 0. \] Find an invertible linear operator \(U\) on \(\RR^2\) such that \[ (U^{\dagger}q)(x_1, x_2) = ax_1^2 + \left(c - \frac{b^2}{a}\right)x_2^2. \] (Hint: To find \(U^{-1}\) (and hence \(U\)), complete the square. For the definition of \(U^{\dagger}\), see part (e) of Exercise 4.)
Solution
1. Complete the square in the original variables.
Because \(a\ne0\), \[ \begin{aligned} q(x_1,x_2) &=a x_1^2+2b x_1x_2+c x_2^2\\ &=a\left(x_1+\frac ba x_2\right)^2 +\left(c-\frac{b^2}{a}\right)x_2^2. \end{aligned} \] Thus the coordinates \[ y_1=x_1+\frac ba x_2, \qquad y_2=x_2 \] diagonalize the quadratic form.
2. Invert the coordinate change to obtain \(U\).
Solving the equations in step 1 for \((x_1,x_2)\) gives \[ x_1=y_1-\frac ba y_2, \qquad x_2=y_2. \] Hence define \[ U(y_1,y_2) =\left(y_1-\frac ba y_2,\ y_2\right), \] so that \[ [U]= \begin{pmatrix} 1&-b/a\\ 0&1 \end{pmatrix}. \] Its determinant is \(1\), so \(U\) is invertible.
3. Verify the requested pullback formula.
By definition, \[ (U^\dagger q)(x_1,x_2)=q(U(x_1,x_2)). \] Substituting the formula from step 2, \[ \begin{aligned} q\left(x_1-\frac ba x_2,x_2\right) ={}&a\left(x_1-\frac ba x_2\right)^2 +2b\left(x_1-\frac ba x_2\right)x_2 +cx_2^2\\ ={}&a x_1^2 +\left(c-\frac{b^2}{a}\right)x_2^2. \end{aligned} \] Therefore \[ \boxed{ (U^\dagger q)(x_1,x_2) =a x_1^2+\left(c-\frac{b^2}{a}\right)x_2^2.} \]