"Deformation" of the kernel of a linear map 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?
 A: 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.)
A: Yes. Choose $x_1,\ldots,x_{m-l}\in \mathbb R^n$ such that $y_1(t)=A(t)x_1,\ldots,y_{m-l}(t)=A(t)x_{m-l}$ form a basis for $\mathrm{img}(A(t_0))$ when $t=t_0$, and $y_{m-l+1},\ldots,y_{m}\in \mathbb R^m$ which extend $y_1(t_0),\ldots,y_{m-l}(t_0)$ to a basis for $\mathbb R^m$. Then the determinant
$$\det\begin{pmatrix} y_1(t) | \cdots | y_{m-l}(t) | y_{m-l+1} | \cdots | y_m\end{pmatrix}$$
varies smoothly with $t$, so we have some neighborhood $\tilde U\subseteq U$ of $t_0$ on which it is nonzero. Thus on $\tilde U$ the function
$$Q(t)=P\begin{pmatrix} y_1(t) | \cdots | y_{m-l}(t) | y_{m-l+1} | \cdots | y_m\end{pmatrix}^{-1}$$
is smooth, where $P$ denotes projection onto the first $m-l$ coordinates. Note that $Q(t)$ projects $\mathrm{img}(A(t))$ onto $\mathbb R^{m-l}$.
Define $a:\tilde U\times \mathbb R^n\to \mathbb R^{m-l}$ by $a(t,x)=Q(t)A(t)x$. Rephrasing your question, you want to know whether $\cup_{t\in \tilde U} \ker(A(t))=a^{-1}(0)$ is a smooth manifold. To see this, note that any smooth local parametrization of $a^{-1}(0)$ near $\{t_0\}\times \{0\}$ extends to a smooth parametrization for some neighborhood of $\{t_0\}\times \ker (A(t_0))$ by linearity, and so we have some neighborhood $\tilde U'$ of $t_0$ such that $f(t,\mathbb R^l)=\ker(A(t))$ for $t\in \tilde U'$.  By a standard theorem, it suffices to check that $0$ is a regular value of $a$. But $a$ is a surjective linear map for all fixed $t$, so its differential at any point must be surjective, thus all values of $a$ are regular.
