Solving a system of non-linear equations with 10 equations and 10 unknowns 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 $\phi_{ab}=\phi_{ba}$.
 A: With the updated question...
Given the comments, I would write the equation using matrices. We are searching for $X\in\mathbb R^{2\times 5}$, such that
$$ X^T X = \Phi, $$
where $\Phi$ is a given symmetric matrix with ones on the diagonal. 
Notice that the equation has a solution if only if $\Phi$ is positive semidefinite with rank $\le 2$. In that case we can compute $X$ using the eigenvalue decomposition of $\Phi$:
$$ \Phi = UDU^T, $$
where $D$ is a diagonal matrix with the eigenvalues, say $d_1 \ge \dotsb \ge d_5$, on the diagonal and $U$ is a matrix with the eigenvectors as columns.
Case 1 $\Phi$ has rank $1$: 
That is $d_1 > 0$ and $d_2\dotsb=d_5=0$. 
Then, choose the first row of $X$ as $\sqrt{d_1}$ times the first column of $U$.
Case 2 $\Phi$ has rank $2$: 
That is $d_1 \le d_2 > 0$ and $d_3\dotsb=d_5=0$. 
Then, choose the first row of $X$ as $\sqrt{d_1}$ times the first column of $U$, and choose the second row of $X$ as $\sqrt{d_2}$ times the second column of $U$.
Notes: If you compute the eigenvalue decomposition numerically, $\Phi$ might appear to have a higher rank than it actually has, due to rounding error and approximation.
A: A Groebner basis computation with Macaulay2 gives
$$0
=
+\phi_{ab}\phi_{ac}\phi_{bd}\phi_{ce}\phi_{de}-\phi_{ab}\phi_{ac}\phi_{be}\phi_{cd}\phi_{de}\\
+\phi_{ab}\phi_{ad}\phi_{be}\phi_{cd}\phi_{ce}-\phi_{ab}\phi_{ad}\phi_{bc}\phi_{ce}\phi_{de}\\
+\phi_{ab}\phi_{ae}\phi_{bc}\phi_{cd}\phi_{de}-\phi_{ab}\phi_{ae}\phi_{bd}\phi_{cd}\phi_{ce}\\
+\phi_{ac}\phi_{ad}\phi_{bc}\phi_{be}\phi_{de}-\phi_{ac}\phi_{ad}\phi_{bd}\phi_{be}\phi_{ce}\\
+\phi_{ac}\phi_{ae}\phi_{bd}\phi_{be}\phi_{cd}-\phi_{ac}\phi_{ae}\phi_{bc}\phi_{bd}\phi_{de}\\
+\phi_{ad}\phi_{ae}\phi_{bc}\phi_{bd}\phi_{ce}-\phi_{ad}\phi_{ae}\phi_{bc}\phi_{be}\phi_{cd}\\
\Longleftrightarrow\\
0 =
+\phi_{ab}\phi_{ac}\phi_{de} (\phi_{bd}\phi_{ce}-\phi_{be}\phi_{cd})\\
+\phi_{ab}\phi_{ad}\phi_{ce} (\phi_{be}\phi_{cd}-\phi_{bc}\phi_{de})\\
+\phi_{ab}\phi_{ae}\phi_{cd} (\phi_{bc}\phi_{de}-\phi_{bd}\phi_{ce})\\
+\phi_{ac}\phi_{ad}\phi_{be} (\phi_{bc}\phi_{de}-\phi_{bd}\phi_{ce})\\
+\phi_{ac}\phi_{ae}\phi_{bd} (\phi_{be}\phi_{cd}-\phi_{bc}\phi_{de})\\
+\phi_{ad}\phi_{ae}\phi_{bc} (\phi_{bd}\phi_{ce}-\phi_{be}\phi_{cd})\\
\Longleftrightarrow\\
0 =
+\phi_{ab}\phi_{ac}\phi_{de} (\phi_{bd}\phi_{ce}-\phi_{be}\phi_{cd})\\
+\phi_{ad}\phi_{ae}\phi_{bc} (\phi_{bd}\phi_{ce}-\phi_{be}\phi_{cd})\\
+\phi_{ab}\phi_{ad}\phi_{ce} (\phi_{be}\phi_{cd}-\phi_{bc}\phi_{de})\\
+\phi_{ac}\phi_{ae}\phi_{bd} (\phi_{be}\phi_{cd}-\phi_{bc}\phi_{de})\\
+\phi_{ab}\phi_{ae}\phi_{cd} (\phi_{bc}\phi_{de}-\phi_{bd}\phi_{ce})\\
+\phi_{ac}\phi_{ad}\phi_{be} (\phi_{bc}\phi_{de}-\phi_{bd}\phi_{ce})\\
\Longleftrightarrow\\
0 =
+(\phi_{ab}\phi_{ac}\phi_{de}+\phi_{ad}\phi_{ae}\phi_{bc})
 (\phi_{bd}\phi_{ce}-\phi_{be}\phi_{cd})\\
+(\phi_{ab}\phi_{ad}\phi_{ce}+\phi_{ac}\phi_{ae}\phi_{bd})
 (\phi_{be}\phi_{cd}-\phi_{bc}\phi_{de})\\
+(\phi_{ab}\phi_{ae}\phi_{cd}+\phi_{ac}\phi_{ad}\phi_{be})
 (\phi_{bc}\phi_{de}-\phi_{bd}\phi_{ce})
$$
Perhaps someone finds a nicer representation.
If one assumes that the vectors
$a=(a_1,a_2),b=(b_1,b_2),c=(c_1,c_2),d=(d_1,d_2),e=(e_1,e_2)$
are the corners of a regular pentagon a Groebner basis computation gives
$$
\phi_{ad}=\phi_{be},\\
\phi_{ac}=\phi_{bd}=\phi_{ce},\\
\phi_{ab}=\phi_{bc}=\phi_{cd}=\phi_{de}
$$
With the given additional info on the symmetry of the $\phi_{\cdot,\cdot}$ it follows from the cyclic symmetry of the 10 equations that
$$
\phi_{ae}=\phi_{ba}=\phi_{ab},\\
\phi_{ad}=\phi_{be}=\phi_{ca}=\phi_{ac},\\
\phi_{ac}=\phi_{bd}=\phi_{ce}=\phi_{da}=\phi_{ad}
$$
Thus in the case of a regular pentagon all $\phi_{\cdot,\cdot} = \phi$ for a single given $\phi$.
A: (This is actually a comment)
We are given $10$ equations for $5$ vectors ${\bf a}$, $\ldots$, ${\bf e}\in{\mathbb R}^2$. For any solution $({\bf a}_0,\ldots,{\bf e}_0)$, rotating the five vectors by the same angle $\theta\in{\mathbb R}$ produces a new solution $({\bf a}_\theta,\ldots,{\bf e}_\theta)$. It follows that the $10$ equations are not "independent", and that the data $\phi_{ik}$ have to satisfy a certain a-priori condition. Any strategem to solve the given system of equations has to reconcile this fact.
A: 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


*

*It is positive semidefinite, so every principal minor is nonnegative.

*Since these are vectors in $\mathbb R^2$, any three are linearly dependent, so every $3 \times 3$  principal minor is $0$.


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:


*

*Start with any vector $A$ with $A \cdot A = G_{11}$.

*Take any vector $B$ with $A . B = \phi_{ab}$ and $B \cdot B = G_{22}$ (generically there will be two possibilities).

*Similarly $C$, $D$, $E$ are determined by their dot products with themselves and the previous vectors (if $A$ and $B$ are linearly independent, the dot products with $A$ and $B$ are enough to determine them uniquely).


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 $E$.
