Ley $y$ be an $n \times 1$ vector and $X$ be an $n \times k$ matrix, and that $y$ and $X$ have the following relationship \begin{equation} y = c + AXb \end{equation} where $c$ is $n \times 1$, $A$ is $n \times n$ and $b$ is $k \times 1$. Assume that $A \neq I_n$, $A \neq 0$, $b \neq 0$ and $X$ has full column rank.
Under this setup, I want to show that $[y,X]$ has full column rank.
When $c=0$, $[y,X]$ should have full rank since $A \neq I$ and $y$ cannot be expressed as a linear combination of $X$.
When $c \neq 0$, I want to prove that there does not exist a vector $d$ such that \begin{equation} c + AXb =Xd \end{equation} Although such d should not exist, I cannot prove it... Any help or alternative proof would be much appreciated!
The vector $c$ is unrestricted. Therefore you obtain a counterexample if you define $c$ by the relationship
$$c + AXb =Xd$$
for some $d\neq 0$.
Of course if you simulate such a model with $X$ independent from $c,A,b$ then you should have full rank "almost surely".