Question. Let $n<m$ be positive integers and $A\in\mathbb R^{m\times n}$, $\mbox{rk}(A)=n$. Does there exist $B\in\mathbb R^{n\times(m-n)}$ such that $(I_n\;B)\cdot A$ is invertible? Equivalently, does $A$ have a left inverse $(C\;D)$ with $C\in\mbox{GL}(n,\mathbb R)$, $D\in\mathbb R^{n\times(m-n)}$?
Motivation. I needed/wanted an argument of this type when dealing with diffeomorphisms between manifolds in $\mathbb R^m$, where $A$ is the Jacobian of the composition of a diffeomorphism of $\mathbb R^m$ and an immersion of rank $n$, and $(I_n\;B)$ is a 'projection map'.
Ideas. $\det((I_n\;B)\cdot A)$ is a polynomial in the entries $(b_{11},\ldots)$ of $B$, so if it is zero for all $B$, it is singular over $\mathbb R[b_{11},\ldots]$. By expressing that all coefficients are $0$, I concluded the case $(m,n)=(3,2)$ but it's not obvious how to generalize this approach.
I'd be interested in a constructive proof as well.
Since $rank A= n$, it follows that $A^T A$ is invertible. Since $A^T$ is $n\times m$ matrix, write it as $A^T=(C \ \ D)$ where $C$ is $n\times n$ and $D$ is $n \times (m-n)$ matrices.
Now, consider $(C + It \ \ D)$. Then define a polynomial $p(t)$ by $$ p(t) = \det \left[ ( C + It \ \ D)\cdot A\right]. $$
Since $A^T A$ is invertible, $p(0) \neq 0$.
Choose $t$ such that $\det (C+ It) \neq 0$ and $p(t)\neq 0$. Then the matrix $$ (C+It)^{-1} (C+It \ \ D) = (I \ \ B) $$ is the matrix you want.