# Regularity for the roots of (characteristic) polynomials with given multiplicity

A classical result states that roots of a polynomial are continuous functions of its coefficients. This is, for exemple, a direct consequence of Rouché's theorem.

Using the implicit function theorem, one can further show that the mapping "coefficients $\mapsto$ roots" is smooth, provided one restricts to polynomials having $\textit{simple}$ roots. It is however easy to check this mapping is not $C^1$ near polynomials where two distinct roots merge to one double root (square root behavior).

I'm considering the following problem: Fix any ordering $\prec$ on $\mathbb C$ and order the zeros of a polynomial $P$ of degree $n$ as $x_1\prec\ldots \prec x_n$ (or restrict to the set of polynomials having only real roots if you prefer), and consider the set of polynomials having smaller root $y_1$ of given multiplicity $m_1$, second root of multiplicity $m_2$, etc. Thus, for some $k\geq 1$, $$\{x_1,\ldots,x_n\}=\{\underbrace{y_1,\ldots,y_1}_{m_1},\ldots,\underbrace{y_k,\ldots,y_k}_{m_k}\},\qquad y_1\prec \ldots\prec y_k.$$ I would believe that the mapping "coefficients $\mapsto$ roots" is smooth (say at least $C^1$) once restricted to such a subclass of polynomials, for any fixed multiplicities $m_1,\ldots,m_k$, but I haven't been able to locate such a result in the literature, nor to prove it using similar strategies as in the previous statements. Any idea?

Edit: If you prefer matrices, I am looking for a statement like: The spectrum of Hermitian (resp. normal) matrices is smooth in the matrix coefficients provided we restrict to Hermitian (resp. normal) matrices having eigenvalues with prescribed multiplicities $m_1,\ldots,m_k$.

• You should define your map carefully, I mean what is its domain. Coefficients of polynomials of your class are far from being arbitrary. Jun 12, 2017 at 18:51
• @AlexandreEremenko Maybe you're right but I don't think it is necessary: for instance when studying this map over polynomial having simple roots and showing it is smooth, one never makes explicit the class of coefficients leading to such polynomials. Jun 12, 2017 at 22:14
• Polynomials arising as characteristic polynomial of hermitian matrices (which have real roots) for example cannot give rise to root singularities. So there can be some structure..
– lcv
Jun 13, 2017 at 7:46
• @lcv Oh, this sounds interesting, do you have a reference for that? Thanks. Jun 13, 2017 at 9:22
• @lcv NB: If by root singularity you mean the mapping is smooth then I disagree: take the $2\times 2$ diagonal matrix with real entries $a$ and $b$. The characteristic polynomial is $(z-a)(z-b)$ and the roots $a$,$ab$ are not $C^1$ functions of the coefficients $a+b$ and $ab$ when the roots merge. Jun 13, 2017 at 9:31

I think that there is a smooth (or analytic) result of the kind that you are seeking:

Let $M^m$ be a connected smooth (or analytic) manifold, and let $P:M\times\mathbb{R}\to\mathbb{R}$ be a smooth (or analytic) function such that, for each $m_0\in M$, the function $P_{m_0}:\mathbb{C}\to\mathbb{C}$ defined by $P_{m_0}(t) = P(m_0,t)$ is a polynomial map of degree $n$ with exactly $k$ distinct roots (in $\mathbb{C})$ with (necessarily fixed) multiplicities $(n_1,\ldots,n_k)$, where $n=n_1+\cdots+n_k$ Set $$Z = \left\{ (m,t)\ |\ P(m,t) = 0\right\}\subset M\times \mathbb{C}.$$ Then $Z$ is a smooth embedded submanifold of $M\times \mathbb{C}$, and the left projection $\pi:Z\to M$ is a smooth submersive $k$-fold covering space of $Z$. (In particular, the roots of $P(m,t)$ are smooth functions on $M$ when $M$ is simply-connected.)

A sketch of the argument is as follows: Since the number of distinct roots of $P_m$ is constant in $m$, the multiplicities are locally constant, by an application of Rouché's Theorem, and hence are constant (since $M$ is connected). Now, suppose that $\lambda_0\in \mathbb{C}$ is a root of $P_{m_0}$ of multiplicity $\mu$. Then the polynomial $$F_{\mu-1}(m_0,t) = \frac{\partial^{\mu-1} P}{\partial t^{\mu-1}}(m_0,t)$$ has a $\lambda_0$ as a root of multiplicity exactly $1$ at $m_0$, and hence, by the usual implicit function theorem, there is a unique smooth function $\lambda(m)$ on an open neighborhood $U$ of $m=m_0$ such that $F_{\mu-1}(m,\lambda(m)) = 0$ for all $m\in U$ and $\lambda(m_0) = \lambda_0$. By shrinking $U$, we can guarantee that there are no other roots of $F(m,t)$ near $\lambda(m)$, and, again by making $U$ sufficiently small, we can ensure, for $m\in U$, that $P(m,t)=0$ has only one distinct root in some small neighborhood of $\lambda_0$ and that it must be of multiplicity $\mu$. Hence, it must be a (simple) root of $F_{\mu-1}(m,t)$ and hence it must be $\lambda(m)$.

The rest of the claims now follow from the implicit function theorem, but applied to $F_\nu(m,t)$ for suitable $\nu$ at various different points of $Z$.

Note that, of course, the covering space $\pi:Z\to M$ need not be trivial, i.e., roots (of the same multiplicity) can exchange places when you go around a non-contractible loop in $M$.

In the special case of purely real roots (hyperbolic polynomials), this problem was studied intensely in connection with hyperbolic equations. You can find a huge amount of information in this nice survey by Armin Rainer

Edit I have erroneously interpreted the edit. The following only addresses the smoothness of the mapping $x \mapsto (\mathrm{eigenvalues})$.

Given the edit, the assertion is true in a more general setting (namely, you don't need to fix the degeneracies) and it is contained in theorem 1.10 of Kato Perturbation theory for linear operators :

Theorem (Kato) Let $T(x)$ be an $n\times n$ matrix depending analytically on $x$ in a domain $D_0$ of the complex plane. Let $x_ \in D_0$ (note that $x_0$ may be an exceptional point, i.e., there may be degenerate eigenvalues) and let there exist a sequence $x_n$ converging to $x_0$ such that $T(x_n)$ is normal for $n=1,2,\ldots$. Then all the eigenvalues $\lambda_j$ (and eigenprojectors $P_j$) are holomorphic at $x=x_0$ and there are no nilpotent terms in the spectral decomposition.

The proof essentially uses the fact that, if there was a singularity (which must be of the form $x^{(1/p)}+\ldots$ for some $p$), then the norm of the eigenprojectors would be unbounded, but for normal operators that's impossible.

• I see, Puiseux's theorem plus using the normality assumption, thanks for the explanations. I'm still puzzled about seing how your statement implies the claim "the map (matrix coefficients)$\mapsto$ (eigenvalues) is $C^1$". Am I missing something obvious? Jun 13, 2017 at 23:29