Parametrizing eigenvectors

Question

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.

Answer 1

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.