Can we continuously choose a basis for the kernel of a continuously family of matrices of constant rank over $\mathbb C$? Suppose $f: \mathbb C \to \mathcal M(n \times n; \mathbb C)$ is a continuous function such that for each $\alpha \in \mathbb C$, $f(\alpha)$ has constant rank $r < n$. The question is whether we can choose a continuous function $g: \mathbb C \to \mathcal M(n \times r; \mathbb C)$ such that $g(\alpha)$ is a basis for the kernel $\ker f(\alpha)$. 
There is a similar question asked here. The domain of the parameter was not specified and there is an answer there by considering the domain of parameter to be $\mathbb R$, i.e., $f: \mathbb R \to \mathcal M(n \times n; \mathbb R)$. I am particularly interested in the case of $\mathbb C$ and could not see directly to adapt the answer. Could someone shed some light on this? Thanks. 
p.s。： I asked the author of the answer but got no reply. And it was also mentioned this can be solved by a result of vector bundles. If possible, I would also be interested in the setup in that language. 
 A: Here is a sketch using $G$-principal bundles : a space $X$ is a $G$-principal bundle if there is a free transitive action $G$  on $X$. The idea is that for such space, we obtain a map $p : X \to Y$ (where $Y = X/G$, which is a manifold), with fibers isomorphic to $G$ but without an explicit bijection. So locally, it's a product with $G$ but not globally. For example, the Klein bottle is a $S^1$-principal bundle over $S^1$ which is not a global product $S^1 \times S^1$.
Claim : a  $G$-bundle is trivial (meaning that $X \cong Y \times G$ and the action is multiplication of the left factor) if and only if there is a section, that is a map $\sigma : Y \to X$ with $p \circ \sigma = id_{X/G}$.
Proposition : There exists a space $BG$ so that isomorphism classes of $G$-bundles over $Y$ are classified by $[Y,BG]$. Here $[A,B]$ is the homotopy classes of map $A \to B$. 
In particular, if $Y$ is contractible and $X$ a $G$-principal bundle over $Y$ there is always a section. 
Back to your problem : 
We take $Y= \Bbb C$ and $f : Y \to R$, where $R$ is the space of matrices of rank $r$.  There is a $\rm{GL}_r$-bundle over $R$, called $E_r$. The fiber is as follows : over $A \in R$, the fiber consists of all the bases of $\ker(A)$, abstractly $\rm{GL}(\ker(A))$. This is a $\rm{GL}_r$-principal bundle, and a section of this bundle is exactly the data for each $A \in R$ of a basis of $\ker(A)$. Well, this bundle is certainly not trivial but we don't care : we look at the pullback of $E_r$ by $f$. By definition, fiber of the pullback $f^*E_r$ coincide with fibers of the original bundle, meaning that we have a bundle over $\Bbb C$ so that the fiber over $\alpha$ is given by all basis of $\ker(f(\alpha))$. By the discussion before, a section always exists over $Y$.
Remark : It would be interesting to see an explicit solution, and also to find a non-trivial family with $Y = \Bbb C^*$ instead.
