14
$\begingroup$

If I have a system of linear equations, $A x = c$, with $A$ an $n\times n$ complex matrix, it is relatively easy to see that the set of matrices $A$ for which there is no (complex) solution has measure zero, as this is the set of matrices such that $\det(A) = 0$.

Can something similar be said for systems of quadratic equations?

More precisely, consider a system of $n$ quadratic equations in $n$ variables, which I can always write as $$ \boldsymbol x^\dagger A_i \boldsymbol x + \boldsymbol b_i \cdot \boldsymbol x + c_i = 0, \quad i=1,..., n, $$ where $A_i$ are $n\times n$ complex matrices, $\boldsymbol b_i\in\mathbb C^n$ and $c_i\in\mathbb C$. Does this system have a solution for almost all values of the parameters? In other words, if a given choice of parameters corresponds to no solutions, is it always true that an infinitesimal change of parameters will give me a system which has solutions?

$\endgroup$
5
  • $\begingroup$ "is it always true that an infinitesimal change of parameters will give me a system which has solutions?" It is always true that there exists an infinitesimal change, etc., etc.; it is not always true that every infinitesimal change etc., etc. $\endgroup$ Oct 25, 2017 at 22:07
  • $\begingroup$ @GerryMyerson care to expand a little bit? Isn't this in contrast to Igor's answer? $\endgroup$
    – glS
    Oct 26, 2017 at 11:22
  • $\begingroup$ Even in the linear case, if $A$ is singular, there exist arbitrarily small changes that keep it singular. E.g., multiply it by $1+\epsilon$ for $\epsilon$ arbitrarily small. $\endgroup$ Oct 26, 2017 at 11:34
  • $\begingroup$ @GerryMyerson oh right of course. I meant to say that there exists a small change such that (...), which should be equivalent to say that the set of non-solvable systems has measure zero (in some properly defined metric over the parameters). I'll fix that bit $\endgroup$
    – glS
    Oct 26, 2017 at 11:36
  • 1
    $\begingroup$ Existence of a small change is not equivalent to measure zero. E.g., given any irrational, there's an arbitrarily small change that makes it rational, but the irrationals have full measure. $\endgroup$ Oct 26, 2017 at 11:47

1 Answer 1

16
$\begingroup$

Yes. The magic words are "elimination theory" and "resultant". In essence, the system has a solution unless some determinant (the iterated resultant) vanishes.

$\endgroup$
1
  • $\begingroup$ Can you provide any ore detail on this? Any reference would be useful $\endgroup$
    – LOC
    22 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.