Given a subdomain of GL(n), when is the map from matrices to their matrices of eigenvectors a diffeomorphism? I'm wondering if there are any general conditions on a subdomain of $GL(n)$, which would guarantee that the map from a matrix to its matrix of eigenvectors is a diffeomorphism. 
For example, given a matrix with distinct eigenvalues, does there exist a neighborhood around it such that this map is diffeomorphic? Any results of that nature would be greatly appreciated.
Edit:
For context, there are some results showing for example that if you have a smooth 1-parameter family of matrices, you can choose the eigenvectors such that the matrix of eigenvectors varies smoothly with your parameter. I'm thinking about extensions to these kinds of results to open sets, or perhaps if such an extension is impossible in general.
 A: Whatever the way you define your map, you cannot expect a diffeomorphism for dimension reasons. The simple fact that you can rescale your eigenvectors makes you lose some dimensions (at least n). In a more formal way, you are defining of map from the matrix algebra (over the real or complex numbers I guess) to the quotient of $Gl(n)$ by a subgroup which has a discrete part (reordering) but more importantly here a continuous part.
A: For example, look at the vector space $S(n)$ of symmetric $x\times n$ real matrices and the action of the orthogonal group $O(n)$ by conjugation:
$$
\ell:O(n)\times S(n)\to S(n),\quad \ell(U,X)=UXU^{-1}
$$
Let $\Sigma$ be the linear subspace of diagonal matrices. Then each $O(n)$-orbit hits $\Sigma$ orthogonally, and $\Sigma$ is a slice for each orbit. If the isotropy of $A\in \Sigma$ is trivial (all eigenvalue distinct), the a small neighborhood $S$ of $A$ extends by the $O(n)$ action to a trivial disk bundle along the orbit (the orbit cylinder), and locally any $X$ in the orbit cylinder can be written as $X=UBU^{-1}$ for $B\in S$ smoothly. That means that $X\mapsto (B,U)$ is a diffeomorphism. If the isotropy group is not trivial, the orbit cylinder is an associated bundle, and thus there is no diffeomorphic decomposition.
To see the last statement, consider the the following example (due to Rellich) which rotates a lot:
$$
x_+(t) := \begin{pmatrix} \cos\frac1t \\ \sin\frac1t \end{pmatrix}, \quad
x_-(t) := \begin{pmatrix} -\sin\frac1t \\ \cos\frac1t \end{pmatrix}, \quad
\lambda_\pm(t) = \pm e^{-\frac1{t^2}},
$$
$$
X(t) := (x_+(t),x_-(t))     \begin{pmatrix} \lambda_+(t) & 0 \\
                 0 & \lambda_-(t) \end{pmatrix}
      (x_+(t),x_-(t))^{-1} 
= e^{-\frac1{t^2}}\begin{pmatrix} \cos\frac2t & \sin\frac2t \\
                         \sin\frac2t & -\cos\frac2t 
                                \end{pmatrix}.
$$
Here $t\mapsto X(t)$ and $t\mapsto \lambda_\pm(t)$ are smooth, whereas 
the eigenvectors cannot be chosen continuously.
