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Given a general matrix $A(t), t>0$, with real entries, I would like to know if the eigenvalues of $A(t)$ are continuous functions of $t$. These eigenvalues may be real or complex.

What about the spectral radius?

A classical result from complex analysis states that the roots of a polynomial vary continuously with the coefficients. Can we use the theorem directly to prove the above? or there are other cases where the eigenvalues are actually discontinuous?

What I'm actually doing is trying to prove that there exists a $t$ for which the spectral radius of $A(t)$ is in $(0,1)$, and I'm doing that by proving that the spectral radius is 1 if $t\rightarrow 0$ and 0 if $t\rightarrow \infty$ (which I already know). Then I would invoke the continuity of the spectral radius to say that there must exist a value of $t$ for which the spectral radius is in $(0,1)$

Thank you.

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  • $\begingroup$ Keep in mind that because a matrix has more than one eigenvalue, one must be careful in defining eigenvalue functions $\lambda_1(t), \ldots, \lambda_n(t)$. If one defines these such that $\lambda_1 \le \ldots \le \lambda_n$, then one must be careful when $\lambda_i(t) = \lambda_{i+1}(t)$ for some $t$. $\endgroup$ – Joshua Mundinger Nov 7 '14 at 17:21
  • $\begingroup$ If you know that the maximal root (in absolute value) of a polynomial varies continuously and the characteristic polynomial depends clearly continuously on the matrix, doesn't it answer your spectral radius question? $\endgroup$ – Peter Franek Nov 7 '14 at 17:58
  • $\begingroup$ @Alqatrkapa You're right.. but the transition should be continuous. right? At the least, the spectral radius should be continuous if the eigenvalues are. $\endgroup$ – Moe Nov 7 '14 at 18:50
  • $\begingroup$ @PeterFranek Yeah I guess. From Wikipedia: "The n roots of a polynomial of degree n depend continuously on the coefficients. This result implies that the eigenvalues of a matrix depend continuously on the matrix. A proof can be found in a book of Tyrtyshnikov." It looks like this is true for any matrix... $\endgroup$ – Moe Nov 7 '14 at 18:52
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    $\begingroup$ Presumably, this is assuming that the map $t \mapsto A(t)$ is itself continuous, even though the question doesn't directly say so. $\endgroup$ – Ilmari Karonen Dec 17 '14 at 17:10
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All of these are continuous, since they are the compositions of continuous functions.

The function from the matrix to any coefficient of the polynomial is itself a polynomial on the entries of the matrix, which is continuous. Thus, the function from a matrix to the vector listing the coefficients of the polynomial is continuous. So, the function from a matrix to its characteristic polynomial is continuous.

The function from a characteristic polynomial to its roots is continuous. So, by the continuity of composition, the function from a matrix to its eigenvalues is continuous.

The function $x \mapsto |x|$ is continuous, as is $(x_1,\dots,x_n) \mapsto \max\{x_1,\dots,x_n\}$. So, the function that yields the largest absolute value of an entry of a vector is continuous. So, by composition, the spectral radius function is continuous.

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  • $\begingroup$ If you talk about continuity, you have to specify the topologies. You said: "The function from a characteristic polynomial to its roots is continuous." I guess what you mean here is that the map from the coefficient vector $(a_0,\ldots,a_{n-1})$ to the set of roots of $\sum a_kz^k$ is continous, where on the left you have just the topology of $\mathbb C^n$ and on the right you use the Hausdroff metric. Is that correct? $\endgroup$ – amsmath Apr 7 at 14:48
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Before talking about the continuity of eigenvalues as functions, one must be clear about the term "eigenvalue function". What does it really mean? How is it defined?

If A(t) is an nxn matrix in which each entry is a continuous function of the parameter t, then for each t, A(t) has n eigenvalues (constituting the spectrum of A(t)). The spectrum of A(t) is a multiset and it is uniquely determined for each t (via A(t)). However, how to number (or parameterize) the eigenvalues in the set is tricky and difficult.

It is known that if the domain of t is a disc in the complex plane that contains the origin, then it may be impossible to parameterize the eigenvalues as continuity functions. If t belongs to a real interval or if all the eigenvalues are real, then there exists a selection of n continuous functions that are eigenvalues of A(t) for each t. (See Kato's book Perturbation Theory of Linear Operators.)

Note that eigenvalues are always continuous in the topological sense (i.e. the map from matrices to their spectra is continuous). The roots continuity of polynomials is usually in this sense. The topological continuity and functional continuity of eigenvalues (roots) are related, but not the same.

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