Prove a quadratic form is positive definite

Question

I want to prove - without using eigenvalues- that the quadratic form

$$q(x,y)=Ax^2+2Bxy+Cy^2$$

is positive definite iff $A>0$ and $AC-B^2>0$

This exercise was taken from a practice for a multivariate calculus course, so I'm not sure how to prove it. The exercise suggests completing squares.

Answer 1

You already have a nice answer. Another approach, more from linear algebra: take the form's matrix:

$$\begin{pmatrix}A&B\\B&C\end{pmatrix}$$

Since it is, obviously, a symmetric matrix, it is (semi)positive definite iff all its main minors are (non-negative) positive, and indeed:

First main minor: $\;A>0\;$ (given)

Second main minor (the matrix determinant): $\;AC-B^2>0\;$ (given)

so the matrix (and thus the quadratic form it represents) is positive definite (and, BTW, this means it determines an inner product)

Answer 2

Hint: assuming $A \neq 0$, $$ q(x,y) = Ax^2 + 2Bxy + (B^2/A) y^2 + (C - B^2/A)y^2 = A(x + (B/A)y)^2 + (C - B^2/A)y^2 $$ Under what conditions can we guarantee $q(x,y) > 0$ for all $(x,y) \neq (0,0)$?

If $A = 0$, which non-zero pair $x,y$ yields $q(x,y) = 0$?