Solving a quadratic 9-equation system I need to solve the following system:
$$\begin{cases}
A^TA=B &(1)\\ 
A\vec{x}=\vec{y} &(2)\\ 
\end{cases}
$$
I need $A$, given $B$, $\vec{x}$ and $\vec{y}$.
$A$ and $B$ are both 3-by-3 matrices; $\vec{x}$ and $\vec{y}$ are 3-component vectors. 
B is by construction a positive-definite matrix; $A$ is expected to be close to $I_3$.
In my problem, $B$ is actually the matrix defining the ellipsoid measured during the calibration procedure of a 3D sensor, and $A$ is the inverse of the sensitivity matrix to be found (it is not necessarily symmetric).
So we have 6 independent, quadratic equations in $(1)$ and 3 independent, linear equations in $(2)$ for 9 unknowns (the coefficients of $A$).
Does anyone have a clue about how to proceed to get an analytical solution?
 A: Better than the QR decomposition I suggested above, the SVD decomposition might work. In fact, given
$$
{\bf A} = {\bf U}\,{\bf \Sigma }\;\overline {\bf V} 
$$
the system above becomes
$$
\left\{ \matrix{
  \overline {\bf A} \;{\bf A} = {\bf V}\,{\bf \Sigma }\;\overline {\bf U} {\bf U}\,{\bf \Sigma }\;\overline {\bf V}  = {\bf V}\,{\bf \Sigma }^{\bf 2} \,\overline {\bf V}  = {\bf B}\quad  \Leftrightarrow \quad {\bf B}\;{\rm diagonalizable} \hfill \cr 
  {\bf U}\,{\bf \Sigma }\;\overline {\bf V} \,{\bf x} = {\bf y}\quad  \Leftrightarrow \quad \overline {\bf x} \;{\bf V}\,{\bf \Sigma }^{\bf 2} \,\overline {\bf V} \,{\bf x} = \overline {\bf x} \;{\bf B}\,{\bf x} = \overline {\bf y} \;{\bf y} \hfill \cr}  \right.
$$
so that B,x,y are not totally independent (for real solutions to exist).
Therefrom ${\bf \Sigma }$ and ${\bf V}$ can be obtained from the normalized eigenvalue decomposition of ${\bf B}$.
${\bf U}$ is the rotation matrix that brings ${\bf \Sigma }\;\overline {\bf V} \,{\bf x}$ in ${\bf y}$.
A: Hint
If you multiply the second equation by $A^T$ you get $A^T y = Bx$. One more multiplication by $A$ yields $A Bx = B y$. Now you can solve 9 by 9 linear sistem and check if the quadratic part is satisfied.
