Proving that the matrix is positive definite

Question

I have looked at similar questions under 'Questions that may already have your answer" and unless I have missed it, I cannot find a similar question.

I am trying to answer the following:

Let $A = \left(\begin{matrix} a & b \\ b & d \\ \end{matrix}\right)$ be a symmetric 2 x 2 matrix. Prove that $A$ is positive definite if and only if $a > 0$ and $\det(A) > 0$. [Hint: $ax^2 + 2bxy + dy^2 = a\left(x+\frac{b}{a}y\right)^2 + \left(d-\frac{b^2}{a}\right)y^2$.]

I can see that in order for $\det(A)$ to be greater than $0$, $ad > b^2$. I have also tried to find the eigenvalues corresponding to the standard matrix $A$ (of the quadratic form) and somehow, I get to $ad < b^2$, which is a contradiction. Could someone point me in the right direction?

Answer 1

Hint: $A$ is positive definite iff $\vec x^{T}A \vec x>0$ for all $\vec x = (x, y) \in\mathbb{R^{2}}\setminus \{(0,0\}$

iff $ax^{2}+2bxy+dy^{2}>0$ for all $x,y\in\mathbb{R}$, $(x, y) \neq (0,0)$

iff $a(x+\frac{by}{a})^{2}+(d-\frac{b^{2}}{a})y^{2}>0$ for all $x,y\in\mathbb{R}$, $(x, y) \neq (0,0)$

iff $a\left((x+\frac{by}{2})^{2}+det(A)y^{2}\right)>0$ for all $x,y\in\mathbb{R}$, $(x, y) \neq (0,0)$

It might be interesting for you to note that this is referred to as Sylvester's criterion.

Answer 2

suppose $\pmatrix{a&b\\b&d} $ is positive definite. we need $$ax^2 + 2bxy + dy^2 > 0 \text{ for all } x, y, x^2 + y^2 \neq 0.$$ taking $x = 1, y = 0$ gives $a > 0.$

taking $x = -b, y = a$ gives $ab^2 -2ab^2+da^2 > 0 \to ad - b^2 > 0$

now, for other directions. suppose $a > 0 \text{ and }ad - b^2 > 0.$ then $$a(ax^2 + 2bxy + dy^2) = (ax+by)^2 + (ad-b^2)y^2 > 0$$ which in turn implies $$ax^2 + 2bxy+dy^2 > 0.$$