kyal

I agree with you completely, in fact I came to a similar conclusion. I think that there is an implicit (or perhaps explicit somewhere I overlooked) assumption when solving linear homogeneous systems namely that the unknowns are somehow "independent" in the sense that picking the value of one does not automatically pick the value of the other. In the case of general quadratic systems involving mixed terms (not the simple case of a diagnal matrix I inquired about initially) it is possible in principle to replace each (degree 2) mixed term with a new variable, resulting in total in a new system of n (n - 1) / 2 variables. However whereas in my simple original case the domain of possible solutions is R^n, in the new system I just constructed the sets of values the variables can take on independent of the coefficient matrix [s_ij] and the right hand side constraint of the system (call it B) is just some (possibly disconnected?) subset of R^n (i.e. because of their interdependence). So the task of find the unique solution w.r.t the coefficient matrix [s_ij] and B involves searching through a subset of R^n with a rather complex (and exponential) structure which definitely cannot be done by a Gaussian Elimination. But that is a very loose reasoning I am giving. Not even sure my reasoning was even correct. In any case I will follow up your suggestion and play a bit

Great! Thank you very much, so that settles it then. It appeared logical to me as well but I was still a bit unsure because it seems that the general case of solving systems of equations involving quadratic forms (degree 2 polynomials) is np-complete (I assume that you are familiar with this term, see e.g. http://mathoverflow.net/questions/153436/can-you-efficiently-solve-a-system-of-quadratic-multivariate-polynomials) so there is something different to it. Just out of curiosity (now that my main issue is solved) then it must be that the mixed terms (which don't exist in my case) must make the difference. The interdependence of these variables is not one that can be expressed in a plain linear homogeneous system? Regards

Can the following system of equations be solved using Gaussian Elimination? [latex]\begin{bmatrix}s_{00} & s_{01} & s_{02} & s_{03}\\s_{10} & s_{11} & s_{12} & s_{13}\\ s_{20} & s_{21} & s_{22} & s_{23}\\ s_{30} & s_{31} & s_{32} & s_{33}\\\end{bmatrix}\begin{bmatrix}x^2_0 \\x^2_1 \\x^2_2 \\x^2_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1\\ 1\\ 1\end{bmatrix}[/latex] If one were to let [latex]w_i = x^2_i, 0 \leq i \leq 3,[/latex] then the above system is (trivially) transformed to [latex]\begin{bmatrix}s_{00} & s_{01} & s_{02} & s_{03}\\s_{10} & s_{11} & s_{12} & s_{13}\\ s_{20} & s_{21} & s_{22} & s_{23}\\ s_{30} & s_{31} & s_{32} & s_{33}\\\end{bmatrix}\begin{bmatrix}w_0 \\w_1 \\w_2 \\w_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1\\ 1\\ 1\end{bmatrix}[/latex] Now assuming that [latex]\begin{bmatrix}w'_0 \\w'_1 \\w'_2 \\w'_3 \end{bmatrix} [/latex] is a unique solution to the above system does that mean that [latex]x_i = \pm\sqrt{w'_i}[/latex] is the zero-dimensional solution set for the original system? Clearly this is only the case when [latex]w'_i \geq 0[/latex] I did try to find articles dealing with systems of quadratic forms that have a diagonal Matrix but I could not find anything relevant. If the above approach is wrong can someone point me in the right direction? Thank you in advance