Let $A$ be a $3\times 3$ matrix defined in the following way: $$A=\begin{bmatrix} a & c & 0\\c & b &-c\\0 & -c & 1-a-b\end{bmatrix}$$
I wish to show that $A=BB^t$ for some matrix $B$. Further suppose that $a,b,c\ge0$ and $abc\ne0$.
Since there is a characterization of positive elements in matrices which says that $A$ is positive iff $A$ can be written as $BB^t$ for some matrix $B$.
Towards that end, for $u=[x,y,z] \in \mathbb{R}^3$, I looked at the inner product $\langle Au,u \rangle$ and found out that $$\langle Au,u \rangle=ax^2+by^2+(1-a-b)z^2+2cxy-2cyz$$
By Hilbert's seventeenth problem, a polynomial of degree $n$ is positive iff it can be written as a sum of $2^n$ positive polynomials. How do I write this polynomial as a sum of such polynomials?
Thanks for the help!!
In order for your matrix to be positive definite, all its leading principal minors must be positive. So in addition to $a > 0$ you need $ab - c^2 > 0$ and $ab + b c^2 - a^2 b - a b^2 - c^2 > 0$.
EDIT: For your matrix to be positive semidefinite, you need $ ab + b c^2 - a^2 b - a b^2 - c^2 \ge 0$ and $a+b-a^2-ab-b^2-2c^2\ge 0$.