11
$\begingroup$

Let $B_1,\ldots,B_s$ be $(s\times s)$ symmetric real matrices and $x=\left(x_1,\ldots,x_s\right)^\prime$ a $(s\times 1)$ vector of unknowns. Is there a way or reference theory for studying the analytical solutions of the system of equations: \begin{eqnarray} x^\prime B_1 x & = & 1\\ &\vdots &\\ x^\prime B_s x & = & 1\\ \end{eqnarray} based on the characteristics of the $B_1,\ldots,B_s$ matrices? Put differently, is there an analytical way to finding the solutions of the previous system of symmetric quadratic equations?

$\endgroup$
3
  • $\begingroup$ This is the same as writing $x'B_1x = 1$ and $x'B'_jx = 0$ for $j > 1$, where $B'_j := B_j - B_1$. The set of solutions of $x'B'_jx = 0$ for $j > 1$ is the intersection of quadratic hypersurfaces, and there is probably a lot of literature on this. $\endgroup$
    – pinaki
    Commented May 13, 2021 at 9:03
  • $\begingroup$ Gröbner bases? Or did you specifically mean something specialised to quadratics? $\endgroup$ Commented May 13, 2021 at 9:05
  • $\begingroup$ Peter, I would prefer something specialised to quadratics. $\endgroup$
    – meie73
    Commented May 13, 2021 at 14:14

3 Answers 3

4
$\begingroup$

There is certainly nothing special for quadratics, as you suggest in your comments.

Indeed, a generic system of polynomial equations can be reduced to a system of quadratic ones by introducing extra variables. For instance, if you have a term $x^3y^2$ in a polynomial system you can introduce the three auxiliary variables $t=y^2, u=x^2, v=tu$, and then you can replace that term with $xv$ (and include those three additional quadratic equations in your system).

$\endgroup$
1
$\begingroup$

When the $(B_i)$ are in general position, there are always $2^s$ complex solutions but we don't have control over the number of real solutions.

Let $n=2^{s-1}$. In generic cases, the system can be decomposed as follows: $\{x_i=Q_i(x_s),i<s,P(x_s)=0\}$ where $Q_i \in \mathbb{R}_{2n-1}[x],P \in\mathbb{R}_{2n}[x]$. The complexity is in the resolution of $P(x)=0$; unfortunately the Galois group of $P$ is $nS_{n}$, and therefore, is not solvable when $s\geq 4$.

Conclusion: there are no analytic formulas giving the required solutions as functions of the $(B_i)$.

$\endgroup$
2
  • $\begingroup$ When the matrices are in general position, one of them can be negative definite, in which case there is no solution at all. $\endgroup$ Commented Nov 11, 2021 at 17:32
  • $\begingroup$ @Denis Serre , as I wrote, $2^s$ concerns the complex solutions. $\endgroup$
    – loup blanc
    Commented Nov 12, 2021 at 9:13
0
$\begingroup$

As Federico explained, systems of quadratic are as general as arbitrary systems of polynomial equations of any degrees. So a general method for solving such systems analytically is out of the question. It is only for special systems of quadratic equations where there is additional algebraic structure that one can solve the system explicitly.

A notable example is the classification of simple Lie algebras. It can be seen as solving for structure constants $f_{ij,k}$ which give the coordinate expression of the Lie bracket $[e_i,e_j]=\sum_k f_{ij,k} e_k$ where the $(e_i)$ is some basis of the Lie algebra. The quadratic equations are given by imposing the Jacobi identity. Something similar is done in conformal quantum field theory where the $f_{ij,k}$ become OPE coefficients and the quadratic equations are given by the so-called crossing equation, a kind of associativity statement.

Another remarkable example of quadratic system that can be solved is the one considered in my article with Chipalkatti "Quadratic involutions on binary forms" where we find all $2^s$ solutions predicted by Bezout's Theorem. The system is given on pages 6-7 in terms of rather complicated coefficients $\omega$ defined on page 35. The complete set of solutions of the system, the object of Theorem 3.2, is given explicitly on pages 8-9 of the arXiv version.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .