TITLE: Solving a system of non-linear equations with 10 equations and 10 unknowns
QUESTION [9 upvotes]: 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 $\phi_{ab}=\phi_{ba}$.

REPLY [3 votes]: 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 $\phi$.