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 $m_1,\ldots,m_k$.

  • $\begingroup$ You should define your map carefully, I mean what is its domain. Coefficients of polynomials of your class are far from being arbitrary. $\endgroup$ Commented Jun 12, 2017 at 18:51
  • $\begingroup$ @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. $\endgroup$ Commented Jun 12, 2017 at 22:14
  • $\begingroup$ 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.. $\endgroup$
    – lcv
    Commented Jun 13, 2017 at 7:46
  • $\begingroup$ @lcv Oh, this sounds interesting, do you have a reference for that? Thanks. $\endgroup$ Commented Jun 13, 2017 at 9:22
  • $\begingroup$ @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. $\endgroup$ Commented Jun 13, 2017 at 9:31

3 Answers 3


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.

  • $\begingroup$ 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? $\endgroup$ Commented Jun 13, 2017 at 23:29
  • 1
    $\begingroup$ @AdrienHardy ups no sorry, my bad. I had erroneously interpreted your edit $\endgroup$
    – lcv
    Commented Jun 14, 2017 at 20:10

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