Norm on the determinant of the product of non square matrices. Say $A$ is an $n\times m$ matrix, $B$ is an $m\times m$ matrix and $C$ is an $m\times n$ matrix, can one relate the determinant of $ABC$ with the determinant of $B$ and some extra data on $A,C$, say their norms or the determinants of $AC,CA$? Specifically, I am looking to bound $\det(ABC)$ from above?
Any idea or useful formula would be appreciated.
 A: The determinant is trivially zero when $n > m$, so I assume that $n \leq m$. Similarly, if $\operatorname{rank}(B) < n$, then the determinant must be zero.
Probably the most effective way to extract information about $\det(ABC)$ from its components is to use the Cauchy-Binet formula. In particular, we have (in the notation of the linked article)
$$
\det(ABC) = \sum_{S \in \binom{[m]}{n}} \det(A_{[n],S})\det(B_{S,S})\det(C_{S,[n]}).
$$
We cannot bound $\det(B_{S,S})$ using $\det(B)$ alone. However, if $\|M\|$ denotes the spectral norm (maximal singular value) of $M$, we could say the following:
$$
\det(A_{[n],S}) \leq \|A\|^n, \quad \det(C_{S,[n]})\leq \|C\|^n.
$$
Bounding $\det(B_{S,S})$ is a bit more complicated, but one approach is as follows. Each determinant $\det(B_{S,T})$ for each $S,T \subset \binom{[m]}{n}$ corresponds to an entry of the exterior power $\wedge^n B$, and we have
$$
\|\wedge^n B\| = \sigma_1(B)\sigma_2(B) \cdots \sigma_n(B),
$$
where $\sigma_1(B) \geq \sigma_2(B) \geq \cdots$ are the singular values of $B$. Thus, we have $|\det(B_{S,S})| \leq \|\wedge^k B\|$, which has the above form. Putting all this together, we end up with the (fairly coarse) upper bound
$$
|\det(ABC)| \leq \binom mn (\|A\|\cdot\|C\|)^n \cdot \|\wedge^n B\|.
$$
