Suppose $f: \mathbb C \to \mathcal M({n \times n; \mathbb C})$ is a continuous function such that $f(\alpha) \in \mathcal M(n \times n;\mathbb C)$ has constant rank $r$ for each $\alpha \in \mathbb C$.
This answer here provides that we can choose a basis of $\ker f$ continuously. That is, there is a continuous $g : \mathbb C \to \mathcal M(n \times (n-r); \mathbb C)$ such that for each $\alpha \in \mathbb C$, $g(\alpha)$ is a basis of the kernel of $f(\alpha)$.
Let us further suppose for any $n$ distinct $\alpha$'s, $(\alpha_1, \dots, \alpha_n)$, we can always form a basis $\mathcal B$ of $\mathbb C^n$ by choosing one vector from each subspace $\ker f(\alpha_j)$. Let us now define a continuous path $\gamma: [0, 1] \to \mathbb C^n$, where $\gamma(0) = (\alpha_1, \dots, \alpha_n), \gamma(1) = (\beta_1, \dots, \beta_n)$ and $\gamma(t)$ has disctinct components for each $t \in [0,1]$. Can we choose a basis by above manner, $\mathcal B( \gamma (t) )$ such that they change continuously? If not, what kind of condition can guarantee that?
It appears that if $r=n-1$, then this is trivially true by our assumption. When $r < n-1$, it is not clear whether it is possible. If let $J$ be the index of columns in each $\ker f(\alpha_j)$ we choose to form a basis at $\mathcal B(\gamma(0))$, it seems like we need the columns in $J$ remain linearly independent through out the path which is not clear to me.