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I'm looking for a proof in the literature of the following fact: Let $A_t$ be a $C^1$-function of one argument $t \in (a,b)$ taking values in the self-adjoint $N \times N$ matrices. Suppose that for every $t$ the spectrum of $A_t$ is simple. Denote by $\lambda_t$ a continuous parametrization of an eigenvalue of $A_t$. Then there exists a $C^1$ function $\psi_t$ taking values in $\mathbb{C}^N$ such that $$ A_t \psi_t = \lambda_t \psi_t $$ and $\psi_t$ is normalized $\sum_{j=1}^{N} |\psi_t(n)|^2 = 1$.

It is not hard to give a proof, but it is somewhat awkward to write down, that's why I am looking for a reference of this fact, I couldn't find one in Reed--Simon 4 and Kato. Here is a short sketch of my idea how to proof it: $\psi_t$ lives in $S^N$, whose tangent space at the point $\psi_t$ is $T_{\psi_t} S^N \cong \{\psi_t\}^{\perp}$. Next, we have that $A_t - \lambda_t|_{\psi_t ^{\perp}}$ is invertible (here the simplicity of the spectrum is used). Using this we can define $\psi_t$ as the solution of the ODE $$ \dot{\psi_t} = (A_t - \lambda_t|_{\psi_t ^{\perp}})^{-1} (\dot{A}_t - \dot{\lambda}_t) \psi_t. $$ Now, various computations show that everything is well-defined and indeed give the desired solution.

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Doesn't this follows easily from the implicit function theorem? –  Mariano Suárez-Alvarez Aug 25 '10 at 15:12
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@ Helge: Really? I'm pretty sure that you can find everything you need in Kato's book. Note: he usually deals with spectral projectors rather than eigenvectors, for the latter are not uniquely defined in case of multiple eigenvalues. –  Pietro Majer Aug 25 '10 at 15:22
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Pietro: Helge did say that "suppose that for every t the spectrum of $A_t$ is simple." ;) In any event, for computational work with unsymmetric matrices with possibly multiple eigenvalues, it is best to work with the Schur decomposition. –  J. M. Aug 25 '10 at 15:30
    
It is unclear whether you are asking for a reference or for someone to validate your proof. I seem to recall that one of Arnold's ODE books (either "Ordinary differential equations" or "Geometric methods") had a thorough discussion of the dependence of diagonalization of a matrix depending on a parameter. –  Victor Protsak Aug 25 '10 at 18:02
    
@Pietro: You are right. It follows from spectral projectors. I thought it didn't since this approach doesn't work for analytic functions (one runs into a problem with taking absolute values or something like this). Can you post it as an answer so that I can mark the problem as solved. @Mariano: The implicit function theorem concerns the eigenvalue .... –  Helge Aug 25 '10 at 21:40

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up vote 3 down vote accepted

I do not know a reference, but here is an easy argument. Consider the space $\overline{M}$ of pairs $(A,\lambda)$ where $A$ is a (self-adjoint) $N \times N$ matrix and $\lambda$ is a root of the characteristic polynomial of $A$. Over the open subset of $\overline{M}$ lying above the operators with distinct eigenvalues, the projection onto the first factor is clearly a proper smooth map, with finite fibers; in other words, the space $\overline{M}$ is a covering space over this open set. Since everything in sight is differentiable, the result that you want follows from the lifting of paths in the base. You need to choose an eigenvalue for the lifting to be unique, but you seem to know which one to choose!

Btw, you only need the eigenvalue $\lambda$ to be simple for the same argument to work. In this case you can use the Implicit Function Theorem to lift the path, since the assumption on the simplicity of the root imply that the differential at the point is surjective.

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Forgot to say: once you have an eigenvalue parameterized, you can find the eigenvector by using Cramer's rule, for instance! –  damiano Aug 25 '10 at 15:19
    
Ugh, Cramer. :D Just arbitrarily ignore one of the $n$ implied linear equations, set the corresponding component to an arbitrary value (in inverse iteration routines 1 is the traditional "arbitrary value"), and eliminate as usual. –  J. M. Aug 25 '10 at 15:58
    
There are of course various ways to proceed: I just wanted something that was "obviously as smooth as you can possibly hope for" and for which you can give an explicit formula. If you wanted to implement it, you would probably not use Cramer's rule, but I thought that the point of the question was not computational. –  damiano Aug 25 '10 at 16:07
    
No worries damiano, I still voted it up. :) –  J. M. Aug 25 '10 at 16:16

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