Let $V$ be a vector space equipped with the inner product $<-,->$. Let $A$ and $B$ be marix expressions for $<-,->$ in two different basis for $V$ that are equivalent, so that there exists a nonsingular matrix $P$ s.t. $A=PBP^T$. Then we have that $det(A)=det(P)^2det(B)$. Thus the determinant of the matrix representing $<-,->$ changes when we perform a change of basis on $V$. However, it changes by a square of the underlying field $K$, i.e. by $det(P)^2$.
If we allow for the absorption of nonzero squares when performing a base change then we can speak of the determinant of the matrix representation as an invariant of the form. We call this invariant the discriminant of the form.
My question is, how does perspective of the discriminant relate to the discriminant $b^2-4ac$ that appears under the radical in the quadratic formula? Thank you.
Best Answer
We will use the standard dot product on the vector space $V = K^2$ as the symmetric bilinear form on $V$ (that is more general than an inner product: you do not need anything like positive-definiteness) and assume $K$ does not have characteristic $2$.
Associate to $f(x) = ax^2 + bx + c$ the quadratic form $$Q(x,y) = ax^2 + bxy + cy^2 = \binom{x}{y}\cdot A\binom{x}{y},$$ where $A=\begin{pmatrix}a&b/2\\b/2&c\end{pmatrix}$. The determinant of $A$ is $ac - (b/2)^2 = -(b^2-4ac)/4$, which up to a nonzero square factor is the negative of the discriminant of $f(x)$.
Taking $a\not= 0$ (you want $f$ to be a quadratic polynomial), we can complete the square: $$ Q(x,y)= a(x+(b/(2a))y)^2 + (c-b^2/(4a))y^2. $$ Therefore by a change of basis in $K^2$, $Q$ has diagonal form $au^2 + (c - b^2/(4a))v^2$. The corresponding matrix is diagonal, with diagonal entries $a$ and $c-b^2/(4a)$, so the determinant of that matrix representation is $ac - b^2/4$, which is the same as before.