Let $A(t)$ be a one-dimensional parametrized family of linear operators on $\mathbb{R}^m$ that has smooth dependence on $t$. Let $V_0\subset \mathbb{R}^n$ be an $n$-dimensional invariant subspace for the matrix $A(0)$, meaning that $A(0)V_0 \subset V_0$. I am interested in the question of whether it is possible to assign for each $t$ an invariant subspace of $A(t)$ that depends on the parameter $t$.
To make this question precise, let $\mathcal{G}_{m,n}$ be the set of all $n$-dimensional linear subspaces of $\mathbb{R}^m$. Equip $\mathcal{G}_{m,n}$ with the metric
$$
d_{\mathcal{G}}(V,W) = \sup_{x\in V} \frac{d(x,W)}{\|x\|},
$$
where $d(x,W)$ is the distance between the vector $x$ and the space $W$,
defined as
$$
d(x,W) = \inf_{y\in W} \|x-y\|.
$$
Now I can ask
Question 1: Does there necessarily exist a continuous function $V:\mathbb{R}\to\mathcal{G_{m,n}}$ such that $V(0) = V_0$ and $V(t)$ is an invariant subspace for the matrix $A(t)$?
Question 2: If so, under what conditions is it possible to choose $V(t)$ to be Lipschitz continuous?
First, if the space $V_0$ is spanned by eigenvectors with separated eigenvalues, then first-order perturbation theory ensures that the eigenvectors are Lipschitz, so certainly the subspaces are Lipschitz. Even if the eigenvalues degenerate, but the eigenvectors remain separated, it should be ok.
The issue seems to be the existence of exceptional points and the possible coalescence of eigenvectors into a Jordan block. But I recently computed an example of a family of matrices where $A(0)$ had a Jordan chain of length 2 (so there is an eigenvector $x$ with $A(0)x=\lambda x$ and a Jordan vector $y$ with $A(0)y = x + \lambda y$), and $V_0 = \mathrm{span}\{x,y\}$. In this example it was still possible to choose $V(t)$ near $0$ in a Lipschitz fashion, despite the fact that the eigenvectors had a singularity at $t=0$. The proof relied on more sophisticated perturbation theory, but now I wonder if there is a more general way to arrive at this result.