Given a system of quadratic equations:
$$\forall i\>\>\>x^{\dagger}M_ix = y_i$$
I want to show that if there exists a solution $x\in \mathbb{C}^n$, there must also exist a solution $\hat{x}$ such that:
$$\forall i\>\>\>\hat{x}^{\dagger}M_i\hat{x} = -y_i$$
Or in other words the system of quadratic equations:
$$\forall i\>\>\>x^{\dagger}(-M_i)x = y_i$$
Under what conditions does this hold? By definition, $M_i$ can't be definite; otherwise, the quadratic forms will always be either positive or negative. Are there any other conditions that are needed to make this true?
In my case, $M_i$ is defined as:
$$M_1 =\begin{bmatrix} 0 & ... & 0 &0 &1 \\ 0 & ... & 0 &0 & 0 \\ 0 & 0 & 0 & 0&0 \\ . & . & . & .& .\\ 0 & 0 &... &0 & 0\end{bmatrix}$$ $$M_2 =\begin{bmatrix} 0 & ... &0 & 1 &0 \\ 0 & ... &0 & 0 & 1 \\ 0 & 0 & 0 & 0&0\\ . & . & . & .& .\\ 0 & 0 &... &0 & 0\end{bmatrix}$$ $$M_3 =\begin{bmatrix} 0 &... & 1 & 0 &0 \\ 0 & ... &0& 1 & 0 \\ 0 & ... &0& 0 & 1\\ . & . & . & .& .\\ 0 & 0 &... & 0&0\end{bmatrix}$$ All the way to $M_{n-1}$: $$M_{n-1}=\begin{bmatrix} 0 &1& 0&... & 0 &0 \\ 0 &0&1 &0&... &0 \\ 0 &0&0 &1&... &0 \\ . & . & . & .& .& .\\ 0 & 0 & 0 & 0&0&1\\ 0 & 0 &... & 0&0&0\end{bmatrix}$$
So $M_i$ are matrices with diagonals of 1s moving from top right corner down as $i$ goes to $n-1$