Math 115. Slides from the Lecture of November 1

This handout contains the slides from the lecture of November 1.

Equivalent Forms have the Same Discriminant

Theorem. If f and g are equivalent forms, then they have the same discriminant. Proof. Write f(x, y) = ax^{2}+bxy+cy^{2} , and let d = b^{2}−4ac. Also let F = [ \begin{matrix}a & b/2 \\ b/2 & c \end{matrix} ]; then

\begin{aligned}[ x y] F [ \begin{matrix}x \\ y \end{matrix} ] & = [ x y] [ \begin{matrix}a & b/2 \\ b/2 & c \end{matrix} ] [ \begin{matrix}x \\ y \end{matrix} ] = [ x y] [ \begin{matrix}ax + by/2 \\ bx/2 + cy \end{matrix} ] \\ & = x(ax + by/2) + y(bx/2 + cy) = ax^{2} + bxy + cy^{2} \end{aligned}

= f(x, y) ,

and

det F = ac − b^{2}/4 = −d/4 .

Also let g(x, y) = a′x^{2}+b′xy+c′y^{2} , let d′ = b^{′2}−4a′c′, and let G = [ \begin{matrix}a′ & b′/2 \\ b′/2 & c \end{matrix} ].

Let M ∈ Γ be a matrix that takes f to g, so that g = f ∘ T_{M} . Then

\begin{aligned}g(x, y) = f(T_{M}(x, y)) & = (M [ \begin{matrix}x \\ y \end{matrix} ])^{t} F (M [ \begin{matrix}x \\ y \end{matrix} ]) \\ & = [ \begin{matrix}x \\ y \end{matrix} ]^{t} M^{t}F M [ \begin{matrix}x \\ y \end{matrix} ] \end{aligned}

= [ x y] (M^{t}F M) [ \begin{matrix}x \\ y \end{matrix} ] ,

Recalling that g(x, y) = [ x y] F [ \begin{matrix}x \\ y \end{matrix} ], we then have

G = M^{T} F M .

However, this takes a proof.

First, we note (from the identity (AB)^{t} = B^{t}A^{t}) that

(M^{t}F M)^{t} = M^{t}F^{t}(M^{t})^{t} = M^{t}F M ,

and therefore M^{t}F M is a symmetric matrix.

Write M^{t}F M = [ \begin{matrix}α & β/2 \\ β/2 & γ \end{matrix} ]. Then

a′x^{2} + b′xy + c′y^{2} = g(x, y) = [ x y] [ \begin{matrix}α & β/2 \\ β/2 & γ \end{matrix} ] [ \begin{matrix}x \\ y \end{matrix} ] = αx^{2} + βxy + γy^{2}

for all x, y ∈ ℤ.

This implies a′ = α, b′ = β, and c′ = γ. Thus

M^{t}F M = G .

Taking determinants of both sides of M^{t}F M = G then gives

−\frac{d′}{4} = det G = det(M^{t}F M)

= (det(M^{t}))(det F )(det M) = 1 · det F · 1

= det F = −\frac{d}{4} ,

and therefore d′ = d. □