7
$\begingroup$

Let $p,q$ be monic polynomials in $\mathbb{C}[t]$ and for $\alpha \in [0,1]$, let $c_\alpha := \alpha p + (1-\alpha)q \in \mathbb{C}[t]$. Since the roots of a polynomial vary continuously with respect to its coefficients, following the fundamental theorem of algebra, the locus $$ L(p,q) := \{ z \in \mathbb{C}\mid c_\alpha(z)=0,~\alpha\in[0,1]\} $$ consists of $n$ (not necessarily distinct) continuous paths.

A priori determination of the endpoints of the paths is ostensiblty diffcult, but experimental evidence suggests the following.

Conjecture. Let $P = \{ \lambda_1,\dots,\lambda_n\}$ and $Q=\{\mu_1,\dots,\mu_n\}$ denote the roots of $p$ and $q$, respectively. For $1 \leq i,j\leq n$, let $d(i,j):= |\lambda_i- \mu_j|$. If $$\text{argmin}_{(i,j)}(d) =\{(k,\ell)\},$$ then there is a path from $\mu_\ell$ to $\lambda_k$.

I am interested in knowing whether the above conjecture is known or if it can be established from what is known from complex analysis and the geometry of polynomials. Note that it is not clear what happens when there is a tie.

Notice that if $p$ and $q$ share a simple root $\lambda$, then there is a (degenerate) path from that root to itself (this corresponds to the case when the distance is zero). The picture below contains a typical example Locus of Roots for $n=5$. generated from the following MATLAB code

p=[1 randn(1,5)+i*randn(1,5)]
q=[1 randn(1,5)+i*randn(1,5)]
hold on
scatter(real(roots(p)),imag(roots(p)),'x','r')
scatter(real(roots(q)),imag(roots(q)),'o','b')
for k=0:.01:1
c=k*p+(1-k)*q;
scatter(real(roots(c)),imag(roots(c)),'.','m');
end

Update: Per Christian's answer below, the conjecture, as stated above, is not true; however, I am still interested in a priori determination of the paths (new question posted).

$\endgroup$
4
  • $\begingroup$ I wonder what will happen to polynomials of different degrees... $\endgroup$ Nov 19, 2017 at 6:26
  • $\begingroup$ @??? Polynomials of different degree can be considered by roots appearing /existing 'at infinity'. $\endgroup$ Nov 19, 2017 at 8:10
  • $\begingroup$ May I suggest that you accept Christian's answer to your original question, and post this variation as a new question (with links at each question to the other one)? $\endgroup$ Jan 30, 2018 at 22:30
  • 1
    $\begingroup$ @GerryMyerson: Done! And thank you for the helpful suggestion. $\endgroup$ Jan 31, 2018 at 19:32

2 Answers 2

12
$\begingroup$

Not true. Take $p=z^2-z$, $q=(z+0.1)^2-0.2^2$. Then the zeros of $p$ are at $0,1$, those of $q$ are at $-0.3,0.1$, so the smallest distance is between $0.1$ and $0$. However, $0.1$ moves towards $1$, not $0$ (as one would in fact expect from a picture).

$\endgroup$
4
  • $\begingroup$ Perhaps it's true with the added assumption that both polynomials contain nonreal roots. $\endgroup$ Aug 2, 2017 at 20:30
  • 5
    $\begingroup$ @PietroPaparella: I don't think that will help, you can do a small perturbation of this example that moves the zeros into the complex plane. $\endgroup$ Aug 2, 2017 at 20:34
  • $\begingroup$ In your example, the roots of both polynomials are collinear -- I'm wondering if excluding this case will make the conjecture true. $\endgroup$ Aug 2, 2017 at 21:04
  • 5
    $\begingroup$ @PietroPaparella: I think this has the same answer as in my previous comment, this property is not essential, and a small perturbation will remove it. $\endgroup$ Aug 2, 2017 at 21:05
4
$\begingroup$

Whatever the truth of the conjecture is (well, Christian's answer tells us...) the zeros of linear combinations of polynomials have been studied, see for example:

On the zeros of a linear combination of polynomials (Pacific Journal, 1966m Robert Vermes).

$\endgroup$
1
  • $\begingroup$ I've read that paper; unfortunately, it does not address the issue I raise above of a priori determination of the paths. $\endgroup$ Aug 3, 2017 at 4:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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