Let \(q\) be the quadratic form on \(\RR^3\) given by \[ q(x_1, x_2, x_3) = x_1x_2 + 2x_1x_3 + x_3^2. \] Find an invertible linear operator \(U\) on \(\RR^3\) such that \[ (U^{\dagger}q)(x_1, x_2, x_3) = x_1^2 - x_2^2 + x_3^2. \] (Hint: Express \(U\) as a product of operators similar to those used in Exercises 5 and 6.)
Solution
1. Rewrite the quadratic form as one square plus a pure cross term.
For variables \(z_1,z_2,z_3\), \[ \begin{aligned} q(z_1,z_2,z_3) &=z_1z_2+2z_1z_3+z_3^2\\ &=(z_3+z_1)^2+z_1(z_2-z_1). \end{aligned} \] Set \[ a=z_1, \qquad b=z_2-z_1. \] Then \[ ab =\left(\frac{a+b}{2}\right)^2 -\left(\frac{a-b}{2}\right)^2. \] Hence the new coordinates \[ y_1=\frac{z_2}{2}, \qquad y_2=z_1-\frac{z_2}{2}, \qquad y_3=z_3+z_1 \] give \[ q(z_1,z_2,z_3)=y_1^2-y_2^2+y_3^2. \]
2. Invert the coordinate change and define \(U\).
Solving the equations in step 1 gives \[ z_1=y_1+y_2, \qquad z_2=2y_1, \qquad z_3=y_3-y_1-y_2. \] Therefore define \[ U(x_1,x_2,x_3) =\bigl(x_1+x_2,\ 2x_1,\ x_3-x_1-x_2\bigr). \] Its standard matrix is \[ [U]= \begin{pmatrix} 1&1&0\\ 2&0&0\\ -1&-1&1 \end{pmatrix}, \] whose determinant is \(-2\). Hence \(U\) is invertible.
3. Verify the required diagonal form.
Let \[ z_1=x_1+x_2, \qquad z_2=2x_1, \qquad z_3=x_3-x_1-x_2. \] Then \[ z_3+z_1=x_3, \qquad z_2-z_1=x_1-x_2. \] Using the decomposition from step 1, \[ \begin{aligned} (U^\dagger q)(x_1,x_2,x_3) &=q(z_1,z_2,z_3)\\ &=(z_3+z_1)^2+z_1(z_2-z_1)\\ &=x_3^2+(x_1+x_2)(x_1-x_2)\\ &=x_1^2-x_2^2+x_3^2. \end{aligned} \] Thus \[ \boxed{(U^\dagger q)(x_1,x_2,x_3)=x_1^2-x_2^2+x_3^2.} \]