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Suppose we have two quadratic equations in $\mathbb R[x_1,\dots, x_n]$. What is the expected dimension of their intersection?

In general what can we say about intersection of $k$ quadratics? How many intersections we expect no real solutions in the best case (which I think should be treated similar to $\mathbb C$ setting)? Is there a way to understand with Bezout`s theorem?

This is what I have. Given a set of solutions ($0$ dimensional points planted) I can generate many equations of degree $2$ that are 'random' looking which have these as common solutions. How many should I generate in any reasonable sense to have only these 0 dimensional points as solutions? I was thinking $O(1)$ should suffice for most purposes (which is best case for me). However may be we need as many as $Ω(n)$ where $n$ is number of variables (which will make my system infeasible)? Is there rigorous analysis we can make?

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First (highlighted) question: the dimension is at most $n-2$. Each equation is expected to decrease the dimension by $1$.

Second question: the dimension of intersection of generic (random) $k$ hypersurfaces is at most $n-k$. (For complex solutions it is equal). It does not matter that they are quadratics.

Third question: Taking $n$ random equations which your points satisfy, you expect that the system will have the set of solutions of dimension $0$, that is finitely many points. It is likely that there will be more points than the number that you assigned. (Besout's theorem predicts that the number of points is the product of degrees, $2^n$ in your case of quadratic equations. So you add one more random quadratic equation which your points satisfy, to obtain your points exactly.

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  • $\begingroup$ So $n+1$ random equations needed even if $1$ planted solution? $\endgroup$
    – Turbo
    Commented May 24, 2019 at 18:02
  • $\begingroup$ @Brout: yes, $n$ will not be enough. $\endgroup$
    – Alexandre Eremenko
    Commented May 24, 2019 at 18:05
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    $\begingroup$ Is this just plain vanilla elimination theory? Would you know the complexity when there are no linear terms and no $x_i^2$ terms (there are only $x_ix_j$ terms and constant term)? $\endgroup$
    – Turbo
    Commented May 24, 2019 at 18:07
  • $\begingroup$ I don't know but if you type "complexity of elimination" on the Google, you immediately get a list of papers on the subject. $\endgroup$
    – Alexandre Eremenko
    Commented May 24, 2019 at 18:12

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