I received the following task as homework.
Given a linear controllable system $\dot{x}=Ax+Bu,\ x\in\mathbb{R}^n,\ u\in\mathbb{R}^n$ derive a criterion of the system being controllable to a manifold $Dx=b$, where $D$ is an $r\times n$ matrix of full rank and $Dx=b$ has at least one solution.
Controllable to a manifold means that for any $x_0\in \mathbb{R}^n$ there exist time T and bounded control $u: [0, T]\to \mathbb{R}^n$ such that under that control $Dx(T)=b$.
I believe I have solved the task. However, the fact that my answer is independent of $b$ bothers me. I would like to get a second opinion.
My attempt:
Define $R=[B,AB,\ldots, A^{n-1}B]$. It can be shown that the span of columns of $R$ (denoted Im $R$) is exactly the space of states reachable from 0. It is also easy to see that for any $y\in\mathbb{R^n}$ and for any $t\in\mathbb{R}$ there exists some $x_0\in\mathbb{R}^n$ such that for some solution of the above system with zero control $x(0)=x_0, x(t)=y$. Basically, to show that any state $x_0$ may be controlled to a solution of $Dx=b$ we need to show that for any $y\in\mathbb{R^n}$ $$y+\text{Im} R\ \cap\ \{x|\ Dx=b\}\neq\emptyset.$$
Let $z$ be any solution of $Dx=b$. Then $\{x|\ Dx=b\}=z+\text{Ker}\ D$. Since $y$ is arbitrary, we have to show that $$y+\text{Im} R\ \cap\ \text{Ker}\ D\neq\emptyset.$$
Equivalently, we need to show that $\dim (\text{Im} R\ + \text{Ker}\ D)=n.$
$$\dim (\text{Im} R\ + \text{Ker}\ D)=\dim (\text{Im} R) + \dim(\text{Ker}\ D)-dim(\text{Im} R\ \cap\ \text{Ker}\ D)=\\ =\text{rank}\ R+n-\text{rank}\ D - \text{rank}\ R + \text{rank}\ DR = n-\text{rank}\ D+\text{rank}\ DR.$$
So, a linear controllable system is controllable to a manifold iff $\text{rank}\ D=\text{rank}\ DR$.
Thanks.