I have an $m$-by-$m$ matrix $A = BCD$ where $B$ is an $m$-by-$n$ matrix, $C$ is an $n$-by-$n$ matrix, and $D$ is an $n$-by-$m$ matrix. My question is: what are the requirements on matrices $B$ and $C$ so that matrix $A$ is always invertible for every invertible matrix $C$? Note that $m < n$ and all matrices are real matrices.
If $m = n$, I can look at the determinant, as $\det(A) = \det(B) \det(C) \det(D)$. To make a matrix invertible, the determinant cannot be zero. Therefore, to make matrix $A$ always invertible for every invertible matrix $C$, we can choose matrices $B$ and $D$ such that $\det(B) \neq 0$ and $\det(D) \neq 0$. However, for the case $m < n$, I don't know where to look.
There is no such condition. When $m<n$, there always exists an invertible matrix $C$ such that $BCD$ is singular.
Pick any nonzero vector $u$. If $Du=0$, then $BCD$ is singular for every $C$ because $BCDu=BC0=0$.
Suppose $Du\ne0$. Since $m<n$, the matrix $B$ has deficient column rank. Hence $Bv=0$ for some nonzero vector $v$. As both $Du$ and $v$ are nonzero, there exists an invertible matrix $C$ such that $v=CDu$. Now $BCD$ is singular because $BCDu=Bv=0$.