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If we have $n-1$ quadratic forms for $n$ variables $x_i$,

$$p_i(x) = M^{(i)}_{jk} x_j x_k$$

for $1\leq i \leq n-1$ and $1 \leq j,k \leq n$ then the zeros of all $p_i(x)$,

$$p_i(x) = 0$$

is generically $0$ dimensional in projective space, i.e. points. I guess generically there are $2^{n-1}$ of such points.

What can be said about this set of points generically? For instance, is there a necessary and sufficient condition on the quadratic forms $M^{(i)}$ such that indeed we have a zero dimensional set of solutions? Is it true that the set of points are roots of a $2^{n-1}$ order polynomial in one variable?

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  • $\begingroup$ What do you mean? Quadratic forms are homogeneous polynomials of degree 2. So when you have a common (non trivial) zero, the quadratic forms all vanish on the whole line generated by this vector. You are probably viewing your problem in projective space right? Then you should make it clear in your question. $\endgroup$ Commented Jan 5, 2022 at 13:30
  • $\begingroup$ Yes, of course, I added this explicitly. $\endgroup$ Commented Jan 5, 2022 at 13:40
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    $\begingroup$ "The set of points are roots of a $2^{n−1}$ order polynomial": what does that mean? A polynomial in how many variables??? $\endgroup$ Commented Jan 5, 2022 at 14:11
  • $\begingroup$ One variable. Edited to make it clear. $\endgroup$ Commented Jan 5, 2022 at 14:16
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    $\begingroup$ @abx: the question is not very well formulated but it does make sense. In fact, this is the claim to fame of Etienne Bezout: finding the degree of the equation resulting from the elimination of all variables except one. $\endgroup$ Commented Jan 5, 2022 at 15:34

2 Answers 2

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To reach a satisfactory understanding of the problem at hand, I think you need to learn about multidimensional resultants (see below for where to get started).

Working over the field $\mathbb{C}$, let $F_1(x),\ldots, F_n(x)$ be $n$ homogeneous polynomials of respective degrees $d_1,\ldots,d_n$. Then there is a polynomial in the coefficients of these forms called the resultant $\mathrm{Res}(F_1,\ldots,F_n)$ which is zero iff there exists $x\in\mathbb{C}^n\backslash\{0\}$ such that $F_1(x)=0,\ldots,F_n(x)=0$.

This resultant is multihomogeneous of degree $\prod_{j\neq i}d_j$ in the coefficients of the form $F_i$.

Now for your situation, one should consider $\mathrm{Res}(p_1,\ldots,p_{n-1},u)$ where $u(x)$ is a generic linear form. If I remember correctly, the common zero set in projective space of your $n-1$ quadratics is zero dimensional iff the above polynomial in the coefficients of $u$ does not vanish identically (with the $p$'s fixed).

Finally, if this zero dimensional condition is satisfied and you look for the coordinates of the solutions of the system along a fixed axis, these indeed are the roots of a degree $2^{n-1}$ polynomial in one variable. This follows from the above polynomial being of degree $2^{n-1}$ with respect to $u$.

A great introductory reference for multidimensional resultants is the book chapter "Introduction to residues and resultants" by Cattani and Dickenstein.

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It turns out this paper has a lot of useful information on resultants (more than enough to answer the original question): Morozov and Shakirov - Analogue of the identity Log Det = Trace Log for resultants.

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  • $\begingroup$ +1 That is a very interesting article but I don't think it is quite suitable as a very first introduction to the subject. I updated my answer with the inclusion of a pedagogical reference. $\endgroup$ Commented Jan 7, 2022 at 14:37

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