Let $A\in\mathcal M_n$ be an $n\times n$ real [symmetric] matrix which depends smoothly on a [finite] set of parameters, $A=A(\xi_1,\ldots,\xi_k)$. We can view it as a smooth function $A:\mathbb R^k\to\mathcal M_n$.

1. What conditions should the matrix $A$ satisfy so that its eigenvalues $\lambda_i(\xi_1,\ldots,\xi_k)$, $i=1,\ldots,n$, depend smoothly on the parameters $\xi_1,\ldots,\xi_k$?

e.g. if the characteristic equation is $\lambda^3-\xi=0$, then the solution $\lambda_1=\sqrt[3] \xi$ is not derivable at $\xi=0$.

2. What additional conditions should the matrix $A$ satisfy so that we can choose a set of eigenvectors $v_i(\xi_1,\ldots,\xi_k)$, $i=1,\ldots,n$, which depend smoothly on the parameters $\xi_1,\ldots,\xi_k$?

Update - important details

  • The domain is simply connected
  • The rank of $A$ can change in the domain
  • The multiplicities of the eigenvalues can change in the domain, they can cross
  • The matrix $A$ is real symmetric
  • $n$ and $k$ are finite

Update 2

  • A relaxation of the conditions of the problem: For fixed $p=(\xi_{01},\ldots,\xi_{0k})$, can we find an open neighborhood of $p$ in the domain and a set of conditions ensuring the smoothness of the eigenvalues and the eigenvectors?
  • $\begingroup$ You want something different from the conditions coming out of the implicit function theorem applied to the characteristic polynomial of $A$ for (1) and the corresponding equation for eigenvalues for (2)? $\endgroup$ – Mariano Suárez-Álvarez Oct 22 '10 at 4:34
  • $\begingroup$ @Mariano Suárez-Alvarez I think that the implicit function theorem helps. But I couldn't see a straight solution. $\endgroup$ – Cristi Stoica Oct 22 '10 at 6:37
  • $\begingroup$ Cristi, you really should have mentioned the "real symmetric" part to begin with; a lot of the worries associated with such problems disappear. $\endgroup$ – J. M. isn't a mathematician Oct 22 '10 at 7:01
  • $\begingroup$ @J. M. you are right, sorry about this. $\endgroup$ – Cristi Stoica Oct 22 '10 at 7:07

The fact that the entries of the matrix are real does seem to help. The state of the art is the following.

  • The spectrum is continuous functions of $\xi$. However, it is not always possible to label the eigenvalues so that they individually are continuous functions.
  • When the multiplicities $m_1,\ldots,m_r$ do not change as $\xi$ varies (no crossing of eigenvalues), then the eigenvalues are as smooth as the matrix. If the domain is simply connected, the eigenvalues may be labelled so as to be smooth functions.
  • When the entries are analytic functions of a single variable ($k=1$) and the eigenvalues remain real, then the eigenvalues may be labelled so as to be analytic functions. However, in case of crossing, this nice labelling is not the obvious one (i.e. not $\lambda_1\le\lambda_1\le\cdots$). This become false for $k\ge 2$, as shown by the example $$\left(\begin{array}{cc} \xi_1 & \xi_2 \\\\ \xi_2 & -\xi_1 \end{array}\right)\qquad\qquad (1).$$
  • The situation is not that good concerning the eigenvectors. The following is called Petrowski's example, $$\left(\begin{array}{ccc} 0 & \xi_1 & \xi_1 \\\\ 0 & 0 & 0 \\\\ \xi_1 & 0 & \xi_2 \end{array}\right).$$ The eigenvalues are real for every $\xi$, distinct when $ \xi_1\ne0$. The matrix is diagonalisable for every $\xi$, but two eigenvectors have the same limit when $ \xi_1\rightarrow0$.

If the domain is not simply connected, you may have additional difficulties with eigenvectors. Take example (1) above, with $\xi$ running over the unit circle $S^1$. When you follow continuously a unit eigenvector $V(\xi)$, it is flipped (i.e. multiplied by $-1$) after one loop

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  • $\begingroup$ I had troubles with the edit of my answer. Thus I deleted it and repost it. Hopefully, it appears only once. If not, please forgive me. $\endgroup$ – Denis Serre Oct 22 '10 at 6:18
  • $\begingroup$ @Denis Serre Thank you. Your answer made me add some details that may be critical, but I omitted. - if the domain is simply connected, can the eigenvalues be labeled as distinct functions, even if their multiplicities change? - if the matrix $A$ is symmetric, can we "save" the eigenvectors? Thank you $\endgroup$ – Cristi Stoica Oct 22 '10 at 7:07
  • $\begingroup$ Note that Denis's first example is a symmetric matrix. An example of a symmetric matrix that shows how the eigenvectors can be very sensitive to changes in matrix entries is due to Wallace Givens: $\begin{pmatrix}1+\epsilon\cos\left(\frac2{\epsilon}\right)&\epsilon\sin\left(\frac2{\epsilon}\right)\\ \epsilon\sin\left(\frac2{\epsilon}\right)&1-\epsilon\cos\left(\frac2{\epsilon}\right)\end{pmatrix}$ $\endgroup$ – J. M. isn't a mathematician Oct 22 '10 at 7:31
  • $\begingroup$ @Cristi. I doubt that it is possible in general. Suppose that there is one complex parameter $z$, and $A(z)$ is holomorphic in $z$. At a branching point, where eigenvalues cross, they form a Riemann surface with non-trivial monodromy. Now, you can write $z=\xi_1+i\xi_2$ and view $A(z)$ as a $(2n)\times(2n)$ matrix, so the trouble applies also to some real matrices. $\endgroup$ – Denis Serre Oct 22 '10 at 8:14
  • $\begingroup$ @J. M. Yes, these are good counterexamples. Maybe there are some conditions under which the eigenvalues and eigenvectors are smooth, at least in the case when the matrix is real and symmetric and the domain simply connected. $\endgroup$ – Cristi Stoica Oct 22 '10 at 8:18

See the following papers by Armin Rainer (et al):

Perturbation theory for normal operators. arXiv:1111.4475

Quasianalytic multiparameter perturbation of polynomials and normal matrices. arXiv:0905.0837.

Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. arXiv:0910.0155

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