# Solving over-determined systems of quadratics in 3 dimensions

I have a 3x3 symmetric matrix $$M$$ with known components $$M_{ij}$$. I'd like to find a vector $$x$$ which has the following relationship with $$M$$

$$M_{ij} = x^T E_{ij} x$$

Where $$E_{ij}$$ is also a 3x3 matrix with known components

$$E_{ij} = \begin{bmatrix} \epsilon_i^0 \epsilon_j^0 & \epsilon_i^0 \epsilon_j^1 & \epsilon_i^0 \epsilon_j^2\\ \epsilon_i^1 \epsilon_j^0 & \epsilon_i^1 \epsilon_j^1 & \epsilon_i^1 \epsilon_j^2\\ \epsilon_i^2 \epsilon_j^0 & \epsilon_i^2 \epsilon_j^1 & \epsilon_i^2 \epsilon_j^2\\ \end{bmatrix}$$

The $$\epsilon_m^n$$ are almost Kronecker delta symbols, i.e. small if $$m \ne n$$, nearly 1 otherwise. In addition, I know the length of the 3 vector $$x$$, so if I break this down, I have 7 equations to solve for 3 unknowns (six from the distinct $$M_{ij}$$, and one $$|x| = r$$).

If I expand those 6 equations out, they yield a generic system of quadratics in 3 variables (all possible terms present), which seems to not have a general solution based on this question.

However, there are a few key differences. My problem has more than double the number of conditions than variables and I have the extra 7th condition for the length, also I'm only interested in the case for 3 dimensions.

If I set the off-diagonal $$\epsilon$$ to 0, I get an approximation that is not good enough unfortunately. I'm currently trying to throw away terms with two small $$\epsilon$$ (i.e. two off diagonals), but haven't had too much luck with that either so far. Does anyone have any ideas about how one could solve this system? I'm thinking there might be some way to get an iterative approximation?

Thanks!

This is very restrictive. Indeed, let $$x=(x_0, x_1, x_2)$$ and $$\epsilon_k=(\epsilon_k^0, \epsilon_k^1, \epsilon_k^2)$$ for each $$k=1,2,3$$. Then for each $$i$$, $$j$$ we have
$$M_{ij}=\sum_{m, n=0}^2 x_mx_n\epsilon^m_i\epsilon^n_j=\left(\sum_{m=0}^2 x_m\epsilon^m_i\right) \left(\sum_{n=0}^2 x_n\epsilon^n_j\right)=(x,\epsilon_i)(x, \epsilon_j) .$$
That is $$M=XX^t$$, where $$X=((x,\epsilon_0), (x,\epsilon_1), (x,\epsilon_2))^t$$. In particular, the matrix $$M$$ should have rank at most $$1$$ . Moreover, for each $$i$$ we should have $$M_{ii}=(x,\epsilon_i)^2\ge 0$$. Then $$(x,\epsilon_i)=\pm\sqrt{ M_{ii}}$$. We should chose the signs assuring $$\operatorname{sign}(x,\epsilon_i)\cdot \operatorname{sign}(x,\epsilon_i)= \operatorname{sign} M_{ij}$$ for each distinct $$i$$ and $$j$$. Satifying the conditions we obtain at most eight possiblities for the matrix $$X$$.
Let $$F$$ be a $$3\times 3$$ matrix whose rows from the top to the bottom are $$\epsilon_0$$, $$\epsilon_1$$, and $$\epsilon_2$$. Then $$Fx=X$$. When $$\epsilon_m^n$$ are sufficiently close to Kronecker $$\delta$$-symbols then the matrix $$F$$ is non-singular by Levy–Desplanques theorem, so then $$x=F^{-1}X$$.
• Thanks, this looks like a good solution. I also think I managed to find a way to solve it by treating each variable ($x^2, y^2, z^2, xy, xz, yz$) as an independent variable and solving it with row reduction like a linear system. So that's another way if someone else comes by one day. Your solution seems like it be more interpretable. – Bobak Hashemi Nov 21 '20 at 0:37