Let $A$ be an $m$ by $n$ $(0,1)$-matrix with $m < n$. If we maximize $\operatorname{det}(AA^T)$ then what property of $A$ are we optimizing? 


This isn't simply maximizing the rank of $A$ and nor is it exactly ensuring that the rows of $A$ are orthogonal.


**July 9 2015**


Fixed typo and changed $A^TA$ to $AA^T$ in question.



What are you maximizing when you maximize the determinant of $A^TA$?
Of course if $A$ is $m$ by $m$ then this is the square of $det (A)$, so the square of the factor by which $A:\mathbb R^m\to \mathbb R^m$ multiplies $m$-dimensional volume.


More generally if $A$ is a linear map from $\mathbb R^m$ to $\mathbb R^n$ with $m\le n$, viewed as a map to an $m$-dimensional subspace of $\mathbb R^n$, then it is still the square of the factor by which $A$ multiplies $m$-dimensional volume.


