I would like to prove the following result:
If $n \geq 2^d$, Quadratic forms $q_1, ... ,q_d : \mathbb{C}^n \rightarrow \mathbb{C}$ have a common zero $x \not = 0, x \in \mathbb{C}^n$
So I know that any such quadratic form w.r.t some basis takes the form $q(x) = \epsilon_1 x_1^2 + ... + \epsilon_m x_m ^2 $ for $\epsilon_i \in \{ \pm 1 \}$ and $ m \leq n$. The two epsilons per term may be related to where the $2^d$ comes from in the result above, but I do not see how exactly.
Furthermore, if I have $q_1$ of this form above in some basis, there is no guarantee $q_2, ... $ will also be diagonalised.
If $q(x) = 0, q( \lambda x) = \lambda^2 q(x) = 0$ so perhaps I can 'count' dimensions somehow and show all the forms must have a common subspace in their kernel, but again I don't see how that would work.
Any help would be appreciated.