Let \(f\) be the form on \(\RR^2\) given by \[ f((x_1, x_2), (y_1, y_2)) = x_1y_1 + 4x_2y_2 + 2x_1y_2 + 2x_2y_1. \] Find an ordered basis in which \(f\) is represented by a diagonal matrix.
Solution
The form factors as \[ f((x_1,x_2),(y_1,y_2)) =(x_1+2x_2)(y_1+2y_2). \] Choose \[ v_1=(1,0), \qquad v_2=(-2,1). \] These vectors form a basis because \[ \det\begin{pmatrix}1&-2\\0&1\end{pmatrix}=1. \] Moreover \[ v_{1,1}+2v_{1,2}=1, \qquad v_{2,1}+2v_{2,2}=0. \] Hence \[ f(v_1,v_1)=1, \qquad f(v_1,v_2)=f(v_2,v_1)=f(v_2,v_2)=0. \] Therefore, in the ordered basis \[ \mathfrak B=((1,0),(-2,1)), \] the matrix of \(f\) is \[ \boxed{[f]_{\mathfrak B}= \begin{pmatrix}1&0\\0&0\end{pmatrix}.} \]