Prove that $q(x,y)=Ax^2+2Bxy+Cy^2$ is positive definite

Question

I want to prove the following:

$q(x,y)=Ax^2+2Bxy+Cy^2$ is positive definite if and only if $A>0$ and $b^2-ac<0$.

I have seen a similar exercise but I want to know why $b^2-ac<0$ must be negative, thanks!

Answer 1

Consider the following simplification:

$$q(x,y) = Ax^2+2Bxy+Cy^2 = A(x^2+\frac{2B}{A}xy)+Cy^2$$

Then, completing the square, we have:

$$x^2+\frac{2B}{A}xy = (x+\frac{B}{A}y)^2-\frac{B^2}{A^2}y^2$$

So, we can rewrite $q(x,y)$ as follows:

$$q(x,y) = A(x+\frac{B}{A}y)^2+Cy^2-\frac{B^2}{A}y^2$$

$$q(x,y) = A(x+\frac{B}{A}y)^2+\frac{CA-B^2}{A}y^2$$

If $A > 0$ and $B^2-AC < 0$, then $A > 0$ and $AC-B^2 > 0$. So, the second term of the expression above is always positive when $y \neq 0$.

In particular, even if the first term may go to $0$, the second term is never $0$ unless $y = 0$. Obviously, this implies that $q(x,y) = 0 \iff x = y = 0$. That proves that the given quadratic form is positive definite under those conditions.