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 \begin{equation} \mathbf{Ay} = \mathbf{1}, \label{eq:linearsolver} \end{equation} where $\mathbf{A} \in \mathbb{R}^{p \times 6}$. If $p \geq 6$, we can obtain $\mathbf{y}$ and further estimate $\mathbf{x}$.

**Questions**:

- Is there any methods besides $\mathbf{Ay}=\mathbf{1}$ to solve the above quadratic equations?
- What is the minimal value of $p$ to guarantee a unique solution of $\mathbf{x}$ of the above quadratic system?

canbe 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. $\endgroup$ – LSpice Jul 21 '20 at 13:23twosolutions is indeed just that the matrix has full rank. $\endgroup$ – lambda Jul 21 '20 at 13:422more comments