How to arrive at these conditions for 2x2 SPD matrices?

Question

Claim: For the matrix $M = \left[ {\begin{array}{cc} a & c \\ c & b \\ \end{array} } \right]$ to be symmetric (trivial) and positive definite: $a>0$ and $ab-c^2>0$ where a,b and c $\epsilon \mathbb{R}$.

My attempt at a proof:

If M is SPD then $z^TMz > 0$ for any real valued 2d vector z.

Let: $z = \left[ {\begin{array}{c} x \\ y \\ \end{array} } \right]$

$z^TMz = ax^2 + 2cxy + by^2$

so

$ax^2 + 2cxy + by^2 > 0$

should give me those conditions,although this is where I am stuck. I can see something that looks kind of right when I rearrange to:

$(\sqrt{a}x+\sqrt{b}y)^2 +2xy(c-\sqrt{a}\sqrt{b}) > 0$

but don't know where to go from there.

I have also tried finding the eigenvalues of M and then forcing them to be > 0, and that did lead to the second condition ($ab-c^2$) but not the first.

What am I not seeing?

Answer 1

Hint: Consider $z$ to be $z := \begin{bmatrix}1 \\ 0 \end{bmatrix}$. For the second property: What do you know about the determinant of an spd matrix?

Answer 2

Look at the 4th characterization here.

$M \in M^{\text{sym}}_{n\times n}(\mathbb R)$ is positive definite if and only if its leading principal minors are positive. The $k$th leading principal minor is the determinant of its upper-left $k\times k$ submatrix.

This theorem is called Sylvester criterion and its proof can be found at the linked page.