# 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!

Update

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:

• I have a feeling you can use the implicit function theorem here to get an idea of what might work. Mar 21 '15 at 20:12
• If the rank$(A(\alpha))=r<n$, then the matrix $A(\alpha)$ has an eigenvalue $\lambda=0$, and the corresponding eigenspace is Null$(A(\alpha)$. While the eigenvalues vary continuously with respect to $\alpha$, it is possible that eigenvectors behave in an unexpected way. Mar 21 '15 at 20:15
• @xecafe, thanks for pointing this. I found some counter-examples when searching for the continuity of eigenvectors of a continuous matrices, but in all of those the discontinuity appears when the dimension of the null space varies. Do you know an explicit counter-example for the case that the dimension of the eigenspace is constant? Mar 21 '15 at 21:19
• @shamisen I don't know any example. Maybe you could find some counter-examples in the book Matrix Perturbation Theory, by Stewart and Sun. Mar 21 '15 at 21:53
• Locally, the answer is clearly yes. Globally, I don't know the answer, but it seems to be negative. See Byrd and Schnabel (1986), Continuity of the null space basis and constrained optimization, Mathematical Programming, 35(1): 32-41. Mar 23 '15 at 9:57

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 $\alpha$.
• Thanks a lot for the answer! Although I understood why you could take block (1,2) of $A(\alpha)$ as $X(\alpha)Y(\alpha)$, I didn't understand why you could take block (2,2) as $Z(\alpha)Y(\alpha)$... could you explain this step in your derivation? Thanks again :) Mar 27 '15 at 18:15
• @shamisen $X(\alpha)$ is invertible. Therefore $\pmatrix{X(\alpha)\\ Z(\alpha)}$ has rank $k$. But $A(\alpha)$ has rank $k$ too. Therefore its last $n-k$ columns are linear combinations of its first $k$ columns. In other words, its last $n-k$ columns are $\pmatrix{X(\alpha)\\ Z(\alpha)}Y(\alpha)$ for some matrix $Y(\alpha)$. Since $A(\alpha)$ is continuous and $X(\alpha)$ is continuous and invertible, it follows that $Y(\alpha)$ is continuous too. Mar 27 '15 at 19:03