Background: Consider the equation $x^T A_1 x = 0$ where $x \in \mathbb{R}^\mu$ and $A_1 \in \mathbb{R}^{\mu \times \mu}$ is a symmetric matrix. Suppose we also demand the normalization $x^T x = 1$. Suppose $A_1$ is indefinite, i.e., suppose $\mu \geq 2$ and there is a positive eigenvalue $\lambda_1$ (corresponding to an eigenvector $v_1$) and a negative eigenvalue $\lambda_2$ (corresponding to an eigenvector $v_2$). Because $A_1$ is symmetric we can assume that $v_1$ is orthogonal to $v_2$. WLOG we can also assume $v_1^T v_1 = 1$ and $v_2^T v_2 = 1$.
Let $x = c_1 v_1 + c_2 v_2$. Then $x^T A x = c_1^2 \lambda_1 + c_2^2 \lambda_2 = 0$ and $x^T x = c_1^2 + c_2^2 = 1$. One can check that this system has solution $$ c_1 = \pm \frac{1}{\sqrt{1 - \frac{\lambda_1}{\lambda_2}}} \qquad c_2 = \pm \frac{1}{\sqrt{1 - \frac{\lambda_2}{\lambda_1}}}. $$ Thus we were guaranteed a solution (and actually were able to write it down) so long as $\mu$ was at least 2 and $A_1$ was indefinite.
Question: Now let us generalize. Suppose we have the following system of equations \begin{align*} x^T A_1 x &= 0 \newline x^T A_2 x &= 0 \end{align*} along with the normalization $x^T x = 1$. Now $x \in \mathbb{R}^\mu$, $A_i \in \mathbb{R}^{\mu \times \mu}$ and both $A_i$ are symmetric. Suppose also that $A_1$ and $A_2$ are algebraically independent (so that we really have two different equations).
If the $A_i$ are indefinite and $\mu = 3$ can we still guarantee a solution like the previous case? Can we go even further and write down a solution? Or perhaps we can add even more assumptions on the $A_i$ to guarantee a solution exists when $\mu = 3$?
Basically I want to find a way to guarantee a (real) solution exists when looking at the zero set of say $\nu$ real quadratic forms using only $\mu = \nu+1$ variables.
The correct theorem is theorem 1 of "Linear Systems of Real Quadratic Forms" by Calabi (1964). The theorem says that if $A_1$ and $A_2$ are real symmetric matrices in $\mathbb{R}^{\mu \times \mu}$ where $\mu \geq 3$ and $$ t_1 A_1 + t_2 A_2 $$ is not positive-definite for all values of $t_1,t_2 \in \mathbb{R}$ then there is a non-trivial solution to the system (but only the case of two quadratic forms).