TITLE: "Deformation" of the kernel of a linear map
QUESTION [5 upvotes]: It is known that the roots of a monic polynomial of fixed degree vary continuously (smoothly?) with its coefficients, at least over $\mathbb{C}$.  My question is whether there is such a result for linear maps.  To be precise:

Let $U \subseteq \mathbb{R}^k$ be an open set and $A: U \to M_{m \times n}(\mathbb{R}) \cong \mathbb{R}^{mn}$ a smooth map.  Suppose that for all $t \in U$, $\ker(A(t))$ is $l$-dimensional.  Fix $t_0 \in U$.  Is there an open set $\tilde{U} \subseteq U$ containing $t_0$ and smooth function $f: \tilde{U} \times \mathbb{R}^l \to \mathbb{R}^n$ such that $f(t,\mathbb{R}^l) = \ker(A(t))$ for all $t \in \tilde{U}$?  

I think this is true.  Here is an example.  Let $A(t) = \begin{pmatrix} \cos t & -\sin t \end{pmatrix}$. Then for $t \in (-\frac{\pi}{2},\frac{\pi}{2})$, we have $\ker A(t) = \langle (\tan t, 1) \rangle$, so we can take $f(t,\lambda) = \lambda (\tan t, 1)$.  Intuitively, it seems that if the rank of a linear map is fixed, nothing else too catastrophic can happen to its level sets - similar to how nothing strange happens with the roots of a polynomial except for their number as the coefficients are varied.  
Can anyone come up with a counterexample, or provide a proof of this fact?

REPLY [4 votes]: Here's a sketch. Working in the subset $W$ of matrices $A$ with rank $r$, we get a continuous map $\rho\colon W\to G(r,n)$ (the Grassmannian of $r$-dimensional subspaces), where $\rho(A)=\text{Row}(A)$. Now, let $C\colon G(r,n)\to G(n-r,n)$ be defined by $C(\xi)=\xi^\perp$. Then $C\circ\rho(A)=\ker(A)$ gives us a continuous (in fact, smooth) map $W\to G(n-r,n)$. ($W$ is in fact a smooth submanifold.)