Let \(q(x_1, x_2) = ax_1^2 + bx_1x_2 + cx_2^2\) be the quadratic form associated with a symmetric bilinear form \(f\) on \(\RR^2\). Show that \(f\) is non-degenerate if and only if \(b^2 - 4ac \neq 0\).
Solution
1. Recover the symmetric matrix of \(f\) from the quadratic form.
If \[ M= \begin{pmatrix} m_{11}&m_{12}\\ m_{12}&m_{22} \end{pmatrix} \] is the standard matrix of \(f\), then \[ q(x_1,x_2) =f(x,x) =m_{11}x_1^2+2m_{12}x_1x_2+m_{22}x_2^2. \] Comparing coefficients with \[ q(x_1,x_2)=a x_1^2+b x_1x_2+c x_2^2 \] gives \[ M= \begin{pmatrix} a&b/2\\ b/2&c \end{pmatrix}. \]
2. Nondegeneracy is equivalent to nonvanishing of the discriminant.
A bilinear form on a finite-dimensional space is non-degenerate exactly when its representing matrix is nonsingular. Here \[ \det M =ac-\frac{b^2}{4} =-\frac14(b^2-4ac). \] Since \(-1/4\ne0\) in \(\mathbb R\), \[ \det M\ne0 \iff b^2-4ac\ne0. \] Therefore \[ \boxed{f\text{ is non-degenerate}\iff b^2-4ac\ne0.} \]