Let $A \in \mathbb{R}^{m \times n}$ and let $\Delta(A)$ be the maximal absolute value of the determinants of the square submatrices of $A$. A simple lower bound would be
$$ \Delta(A) \geq \underset{i,j} \max |A_{i,j}|. $$
I'm wondering what is the best known upper bound on $\Delta(A)$. Does anyone know of such a result? In particular, would orthogonality of $A \in \mathbb{R}^{n \times n}$ help?
A sharp upper bound for the determinant of a square matrix $B_{k\times k}$ is $\det(B)\leq \prod_{i=1}^k\|B_i\|$, where $\|B_i\|$ is the euclidean norm of the column $i$ of $B$.
Assume $B_{k\times k}$ is a submatrix of $A$. The columns of $B$ are "subcolumns" of the columns of $A$. If $A_i$ is the column $i$ of $A$ that contains the column $B_j$ then $\|B_j\|\leq\|A_i\|$. Thus, $\det(B)\leq \prod_{j=1}^k\|A_{i_j}\|$, where $\|A_{i_1}\|\geq\ldots\geq\|A_{i_n}\|$.
Therefore $\Delta(A)\leq\max\{\prod_{j=1}^k\|A_{i_j}\|,\ k=1,\ldots,n$}.
Notice that this bound is attained if $A$ is square and orthogonal.