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Theorem (Rellich). Let $\boldsymbol{A}(t) : \mathbb{R}\rightarrow\mathbb{C}^{n \times n}$ be a Hermitian matrix function that depends on $t$ analytically.

(i) The $n$ roots of the characteristic polynomial of $\boldsymbol{A}(t)$ can be arranged so that each root $\lambda_j(t)$ for $j = 1,\cdots,n$ is an analytic function of $t$.

(ii) There exists an eigenvector $v_j(t)$ associated with $\lambda_j(t)$ for $j = 1,\cdots,n$ satisfying

(1) $(\lambda_j(t)\boldsymbol{I} - \boldsymbol{A}(t))v_j(t)=0\forall t\in\mathbb{R}$

(2) $||v_j(t)||_2=1~\forall t\in\mathbb{R}$

(3) $v_j^*(t)v_k(t)=0~\forall t\in\mathbb{R}$ for $k\neq j$

(4) $v_j(t)$ is an analytic function valued vector of $t$.

My question is what happens if some of the eigenvalues have multiplicity,i.e. $\lambda_j(t)=\lambda_k(t)$ for some values of $t$ if $k\neq j$. I especially need to know whether (i) and (ii)-2 and (ii)-4 remains valid?

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Yes, Rellich's theorem does not require the eigenvalues to be distinct. See e.g. Reed and Simon, "Methods of modern mathematical physics vol. 4: Analysis of Operators", Chapter XII (in particular Problems XII.16 and XII.17).

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  • $\begingroup$ Does Rellich's theorem guarantee global analyticity? (or is it only local) $\endgroup$ – trienko Dec 3 '13 at 14:36
  • $\begingroup$ Can some of the eigenvalues be zero? So the rank is never full rank? $\endgroup$ – trienko Dec 4 '13 at 8:52
  • $\begingroup$ Yes, of course some eigenvalues can be $0$ (e.g. if some row of $A(t)$ is all $0$). $\endgroup$ – Robert Israel Dec 6 '13 at 3:36
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A comment to Robert's answer. Rellich's theorem is actually a simple consequence of the Schwarz symmetry principle from analytic function theory. Take a semi-neighborhood $D$ of the real axis of the shape $0<y<h(x)$, where $h$ is a positive continuous function, so small that there are no singularities of the algebraic function $\lambda(t)$ in $D$. (Algebraic functions have only finitely many singularities, so there is no problem with existence of such $h$). In $D$ the algebraic function breaks into finitely many holomorphic branches. Each branch must have a real limit when $t$ approaches the real axis. Because all eigenvalues of an Hermitean matrix are real. Therefore, by the Schwarz symmetry principle, all branches of $\lambda$ have analytic continuations to a full neighborhood of the real line. The statements about eigenvectors follow from analyticity of $\lambda$.

When some $\lambda_j$ collide on the real line, their graphs simply cross, each remaining analytic. This is called "level crossing" in physics.

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