What is a counterexample to this one?

Question

Let $R$ be a commutative ring and $A\in M_{n\times m}(R)$ where $n\neq m$.

What is an example such that $\det(AA^t)\neq \det(A^tA)$?

Indeed, I think it's true. If this is true, how do I prove this?

Answer 1

How about $R = \mathbb R$ and $A = \begin{bmatrix} 2 & 1 \end{bmatrix}$?

Answer 2

This is not true even in $\mathbf{R}$, try it with any matrix, it will most likely be false. In particular $A={2 \choose 1}$ works.

Answer 3

If $A$ is not square, then the determinants of $A^tA$ and $AA^t$ are equal if and only if $A$ has neither full row rank nor full column rank. In this case the common determinant is $0$

Indeed, if the rank of $A$ is the same as the number of rows, then $AA^t$ is invertible, while $A^tA$ isn't. Similarly in the case the rank is the same as the number of columns.