Principal Minor Theorem for dimension 2

Question

Let $A$ be a real ($2 \times 2$)-matrix such that the map $\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$, $\langle x,y \rangle = x^t A y$ is a scalar product. Now I would like to show that the determinant of $A$ must be positive.

Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. I was able to show that $a,d >0$ and $b=c$, so our matrix has the form $$A = \begin{pmatrix} a & b \\ b & d \end{pmatrix}.$$

The determinant is $ad-b^2$. My idea is it to find a vector $z \in \mathbb{R}^2$ such that $\langle z,z \rangle = \det(A)$ which must be positive due to the positive definiteness of the scalar product.
If $z = \begin{pmatrix} x \\ y \end{pmatrix}$, then $\langle z,z \rangle = ax^2 + 2bxy + dy^2$. How do I find values for $x,y$ so that this is exactly the determinant of $A$? I think I need to use squares of $a,d$ (which is possible since both are positive) or absolute values, but I am not able to find the solution.

Could you please help me with problem? Thank you!

Answer 1

You must have

$$\langle z,z \rangle = ax^2 + 2bxy + dy^2 >0$$ for all $z\neq 0$. In particular for all $z=(x,1)^T$.

In particular the discriminant of the trinomial

$$a x^2 +2bx +d$$ has to be strictly negative. This is exactly what you’re looking for.

Answer 2

You don't have to. If the homogeneous quadratic polynomial $\;ax^2+2bxy+dy^2\;$ takes positive values for all $(x,y)\ne (0,0)$, it means its (reduced) discriminant $\;\delta'=b^2-ad <0$, and this discriminant is also $\;-\begin{vmatrix} a&b\\b&d\end{vmatrix}$.

Answer 3

$$ ax^2 + 2bxy + dy^2 = a(x + \frac{by}a)^2 + \frac1a(ac-b^2)y^2 $$