I'm working on a problem where I seem to have run into a system of non-linear equations. I have ten equations and ten unknowns. In the equations below, all of the $\phi_{ij}$'s are known, but all of the $a_{1},...,e_{2}$ are unknown.
$$ a_{1}b_{1}+a_{2}b_{2}=\phi_{ab}\\ a_{1}c_{1}+a_{2}c_{2}=\phi_{ac}\\ a_{1}d_{1}+a_{2}d_{2}=\phi_{ad}\\ a_{1}e_{1}+a_{2}e_{2}=\phi_{ae}\\ b_{1}c_{1}+b_{2}c_{2}=\phi_{bc}\\ b_{1}d_{1}+b_{2}d_{2}=\phi_{bd}\\ b_{1}e_{1}+b_{2}e_{2}=\phi_{be}\\ c_{1}d_{1}+c_{2}d_{2}=\phi_{cd}\\ c_{1}e_{1}+c_{2}e_{2}=\phi_{ce}\\ d_{1}e_{1}+d_{2}e_{2}=\phi_{de}\\ $$
Would anyone have any strategies for solving this system of non-linear equations for the $a_{1},...,e_{2}$? Is it possible to convert this into a system of linear equations somehow?
Additional info (based on comments):
The dot products of the vectors with themselves are all known quantities, and in fact equal to one: $$ \phi_{aa}=1\\ \phi_{bb}=1\\ \phi_{cc}=1\\ \phi_{dd}=1\\ \phi_{ee}=1\\ $$
Yes, it is true that $\phi_{ab}=\phi_{ba}$.
Considering $(a_1, a_2), \ldots, (e_1, e_2)$ as vectors $A, \ldots, E \in \mathbb R^2$, you're specifying their dot products with each other (but not with themselves), thus all off-diagonal elements of their Gramian matrix. The constraints on the Gramian matrix are
I would start with (2), getting $10$ equations in the $5$ diagonal elements. Elimination is likely to lead to a finite number of possible solutions (if any), and then you can check (1) for each.
Once you have a feasible Gramian matrix $G$, you can find solutions as follows:
EDIT: If the diagonal elements are all known to be $1$, you just have to check that the matrix is positive semidefinite and has rank $\le 2$, and then proceed as above. If all off-diagonal elements are $\pm 1$, then $A$ is any unit vector and all the others are $\pm A$ (depending on whether $\phi_{a\cdot} = +1$ or $-1$). Otherwise, without loss of generality suppose $|\phi_{ab}| < 1$. Let $A$ be any unit vector, and $V$ one of the two unit vectors orthogonal to $A$. Take $B = \phi_{ab} A + \sqrt{1-\phi_{ab}^2} V$. Thus $$V = \dfrac{-\phi_{ab}}{\sqrt{1-\phi_{ab}^2}} A + \dfrac{1}{\sqrt{1-\phi_{ab}^2}} B$$ so that $$ C = \phi_{ac} A + (C \cdot V) V = \phi_{ac} A + \left(\dfrac{\phi_{bc} -\phi_{ab} \phi_{ac}}{\sqrt{1-\phi_{ab}^2}}\right) V$$ and similarly for $D$ and $E$.