suppose we have a system of quadratic equations of the following form \begin{align} \boldsymbol x^T\boldsymbol A_i (\boldsymbol x + \boldsymbol b)=y_i\ ,\quad i=1,\cdots,M \end{align} where $\boldsymbol x\in\mathbb{R}^N$， $\boldsymbol A_i\in\mathbb{R}^{N\times N}$ is the $i$-th given symmetric matrix, $\boldsymbol b\in\mathbb{R}^n$ is a some given vector, $y_i$ is the given $i$-th measurement.
We would like to solve for $\boldsymbol x$. What is the condition under which a solution $\boldsymbol x$ exist for arbitrary $\boldsymbol b$ and $y_i, i=1,\cdots,M$?
There is a similar question posted: https://mathoverflow.net/questions/284368/do-almost-all-systems-of-quadratic-equations-have-solutions But the $\boldsymbol b$ is different for every $i$, and I don't quite understand the answer that uses "elimination theory" and "resultant".
Thanks so much! Any help would be appreciated!
I don't know of an "if and only if", but I can give you some conditions under which solutions will not always exist.
Suppose some linear combination $\sum_i c_i A_i$ (with $c_i$ not all $0$) is positive semidefinite. Then $$\sum_i c_i x^T A_i (x + b) = \sum_{i} c_i \left((x + b/2)^T A_i (x+b/2) - \frac{1}{4} b^T A_i b\right) \ge - \sum_i c_i b^T A_i b/4 $$ is bounded below: thus if $4 \sum_i c_i y_i < - \sum_i c_i b^T A_i b$ there is no solution.