TITLE: What is the relation between $\det(A^TA)$ and $\det(AA^T)$ for an $m\times n$ matrix $A$?
QUESTION [10 upvotes]: Let $A$ be an $m\times n$ real matrix, and $\det(\cdot)$ the determinant. It is rather trivial that when $A$ is of size $1\times 1$, then $\det(A^TA)=\det(AA^T)$. 

In general, what is the relation between $\det(A^TA)$ and $\det(AA^T)$?

REPLY [4 votes]: Case 1: $A$ is a square matrix
If $A$ is a square matrix, then $\det(A^T)=\det(A)$, so
$$
\det(AA^T)=\det(A)\det(A^T)=(\det(A))^2
$$
and the same holds for $\det(A^TA)$.
Case 2: $A$ is not a square matrix
Suppose $A$ is $m\times n$, with $m\ne n$. If $m>n$, then the rank of $AA^T$ is at most $n$ and therefore $\det(AA^T)=0$. If $A$ has rank $n$, then $A^TA$ has rank $n$ (see later), so $\det(A^TA)\ne0$. Similarly if $m<n$.
Thus, the equality $\det(AA^T)=\det(A^TA)$ only holds if and only if the rank of $A$ is strictly less than $\min\{m,n\}$.
Exercise
The rank of $A$ equals the rank of $AA^T$ and of $A^TA$.