How to solve a system of quadratic equations?

Suppose we have a system of $$p$$ quadratic equations about $$\mathbf{x} \in \mathbb{R}^3$$ and $$\mathbf{x} > 0$$

$$\left\{ \begin{array}{lr} \mathbf{x}^\top \mathbf{C}_1 \mathbf{x} = 1, \\ \mathbf{x}^\top \mathbf{C}_2 \mathbf{x} = 1, \\ \quad\quad \vdots\\ \mathbf{x}^\top \mathbf{C}_p \mathbf{x} = 1, \\ \end{array} \right.$$

where matrices $$\mathbf{C}_1, \dotsc, \mathbf{C}_p \in\mathbb{R}^{3 \times 3}$$ are symmetric and positive definite.

Suppose $$\mathbf{x} = [a,b,c]^\top$$, and $$\mathbf{y} = [a^2,b^2,c^2, ab, ac, bc]^\top$$. It is known that the above quadratic equations can be written as $$$$\mathbf{Ay} = \mathbf{1}, \label{eq:linearsolver}$$$$ where $$\mathbf{A} \in \mathbb{R}^{p \times 6}$$. If $$p \geq 6$$, we can obtain $$\mathbf{y}$$ and further estimate $$\mathbf{x}$$.

Questions:

1. Is there any methods besides $$\mathbf{Ay}=\mathbf{1}$$ to solve the above quadratic equations?
2. What is the minimal value of $$p$$ to guarantee a unique solution of $$\mathbf{x}$$ of the above quadratic system?
• What do you mean by "What is the minimal value of $p$ to guarantee a unique solution"? Do you mean the minimal value for which a solution can be unique? No matter how big $p$ is, $\mathbf A$ could be, say, the $0$ matrix, in which case there are no solutions, or some low-rank matrix, in which case there could be multiple solutions. Jul 21 '20 at 13:23
• If $\mathbf x$ is a solution then so is $-\mathbf x$, so you can't ever have uniqueness. Up to sign you can certainly recover $\mathbf x$ from $\mathbf y$ so the condition for having only two solutions is indeed just that the matrix has full rank. Jul 21 '20 at 13:42
• Have you tried in $\mathbb{R}^2$ first? Jul 21 '20 at 15:48
• Related: mathoverflow.net/q/308163 Jul 21 '20 at 16:06
• @LSpice Let's say $\mathbf{A}$ is always full rank in this case. I want to know how many quadratic equations are required to obtain $\mathbf{x} \in \mathbb{R}^3$, is p=3 enough?
– heng
Jul 22 '20 at 0:31