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?

  • $\begingroup$ "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. $\endgroup$ – Gerry Myerson Oct 25 '17 at 22:07
  • $\begingroup$ @GerryMyerson care to expand a little bit? Isn't this in contrast to Igor's answer? $\endgroup$ – glS Oct 26 '17 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. $\endgroup$ – Gerry Myerson Oct 26 '17 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 '17 at 11:36
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Oct 26 '17 at 11:47

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

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