Test case:
Consider the following system of quadratic equations: \begin{equation} \begin{cases} a x^2+2bxy+cy^2=\lambda x\\ bx^2+2cxy+dy^2=\lambda y \end{cases} \end{equation} Where $a,b,c,d\in\mathbb{R}$ and $\lambda\in\mathbb{C}$ are constants.
There are $2^2=4$ possible solutions. How could turn this system of 2 quadratic equations into a set of 4 (coupled) linear equations? That is to say, find a set of constants such that: \begin{equation} \begin{cases} a_1x_1+a_2x_2+a_3y_1+a_4y_2=\lambda x_1\\ b_1x_1+b_2x_2+b_3y_1+b_4y_2=\lambda x_2 \\ c_1x_1+c_2x_2+c_3y_1+c_4y_2=\lambda y_1 \\ d_1x_1+d_2x_2+d_3y_1+d_4y_2=\lambda y_2\\ \end{cases} \end{equation} General Case:
Consider a general system of $N$ quadratic coupled equations where $x\in \mathbb{C}^N$ is the unknown:
\begin{equation} \sum_{jk}^N T_{ijk}x_jx_k=\lambda x_i,\quad \forall 1\leq i \leq N \end{equation} Where $\lambda$ and $T_{ijk}$ are simply constants (and $T$ fully symmetric) to describe these quadratic equations.
Then is it possible to turn this set of $N$ equations, into $2^N$ linear equations? That is to say, find a matrix $M$ of size $2^N\times 2^N$ such that: \begin{equation} \sum_{j}^{2^N}M_{ij}\tilde{x}_j=\lambda \tilde{x}_i,\quad \forall 1\leq i \leq 2^N \end{equation} Such that the solution $\tilde{x}\in\mathbb{C}^{2^N}$ gives the same solution for the quadratic case.