Continuity of the basis of the null space

Consider a real $n\times n$ matrix $A(\alpha)$ continuously dependent of a real vector $\alpha$, i.e. a real matrix-valued continuous function $A: \mathbb{R}^m \rightarrow \mathbb{R}^{n \times n}$.

If the rank of $A(\alpha)$ is the same for every $\alpha \in \mathbb{R}^m$, it is true that it is always possible to continuously choose a basis for the null space of $A(\alpha)$?
That is, there exists a continuous matrix valued function $N(\alpha)$ such that its columns form a basis for $A(\alpha)$ for every $\alpha \in \mathbb{R}^m$?

Thanks for any help!


Just to add more context for this question, for the case that the rank of $A(\alpha)$ is different for some $\alpha$, there are many counterexamples. More information about the non-constant rank case can be seen in the following MO questions:

  1. Conditions for smooth dependence of the eigenvalues and eigenvectors of a matrix on a set of parameters
  2. How to find/define eigenvectors as a continuous function of matrix?

Solutions Collecting From Web of "Continuity of the basis of the null space"

Locally, the answer is clearly yes. Let’s consider $\alpha=0$. Suppose the common rank is $k$. By singular value decomposition or by left and right multiplications of (possibly different) permutation matrices, we may assume that the leading principal $k\times k$ submatrix of $A(0)$ is invertible. Since $A(\alpha)$ is continuous and has rank $k$ for every $\alpha$, it follows that in a neighbourhood of $\alpha=0$, we have
A(\alpha)=\begin{bmatrix}X(\alpha)&X(\alpha)Y(\alpha)\\ Z(\alpha)&Z(\alpha)Y(\alpha)\end{bmatrix}
for some continuous functions $X,Y,Z,W$, with $X(\alpha)$ being a $k\times k$ invertible matrix. So, if $e_1,\ldots,e_{n-k}$ is the standard basis of $\mathbb R^{n-k}$, then
\left\{\begin{bmatrix}-Y(\alpha)e_1\\ e_1\end{bmatrix},\,\ldots,\,\begin{bmatrix}-Y(\alpha)e_{n-k}\\ e_{n-k}\end{bmatrix}\right\}
is a basis of the null space of $A(\alpha)$ and it varies continuously with $\alpha$.