Can a smooth matrix function $M_{ij}(t)$ change rank?

Question

Let $M_{ij}(t)$ be smooth functions of a real parameter $t$, such that $\mathbb M(t) = (M_{ij}(t))$ is a $n\times m$ real matrix. By smooth I mean that $M_{ij}(t)$ is continuous and differentiable.

Suppose $\mathbb M(t)$ is of deficient rank, $r < m,n$ for some interval of values $t \in [t_0,t_1]$. Provided $\mathbb M(t)$ is always smooth in its domain, can can the rank $r(t)$ change?

My puzzlement arises from the fact that $r(t)$, being discrete, will necessarily jump discontinuously. I'm not sure if this can be reconciled with the smoothness of $M_{ij}(t)$. My intuition is that a change of rank implies some "discontinuity" of $M_{ij}(t)$, but maybe at a higher-order derivative?

Answer 1

It is possible for the rank to change. Consider the smooth matrix-valued function $$ \mathbf M:\mathbb R\to {}^1_1 \mathbb R^2 \\ \mathbf{M}(t)=\begin{bmatrix}t & t^2 \\ \sqrt{|t|}& |t|^3\end{bmatrix}$$ $\mathbf M(-1), \mathbf M(1)$ are of rank one and $\mathbf M(0)$ is of rank zero but $\mathbf M(t)$ is of rank two for $t\notin \{-1,0,1\}$.

Answer 2

Some geometric intuition might help here. The space of all matrices is simply $\mathbb R^{m \times n}$, the Euclidean space. A part of it has rank $0$, another part has rank $1$, etc. There's nothing whatsoever that stops you from moving from one part to another, since it's just a plain flat space. The problem is the rank function is discontinuous, so you can't expect any smoothness to carry over.