Problem. Let $A$ be an $m\times n$ matrix and $B$ be an $n\times m$ matrix. Show that $$\det(A\otimes B)=\det(B\otimes A).$$
This formula apparently holds if $A$ and $B$ are square matrices, but here this assumption fails in general.
I am considering the general definition of determinant: $$\det(M)=\sum_{\sigma\in S_n}{(-1)^{\rm sgn(\sigma)}a_{1\sigma(1)}\cdots a_{n\sigma(n)}}.$$
We have $$\det(B\otimes A)=\det((B\otimes A)^T)=\det(B^T\otimes A^T).$$
The shape of $B^T$ is the same as $A$ and the shape of $A^T$ is the same as $B$. Intuitively, the summands in the two determinants are exactly the same, but this argument is vague and complicated. It would be better if anyone has some easier proofs.
Best Answer
When $m\ne n$, both determinants are zero.
Suppose $m<n$. Then $Ax=0$ has a non-trivial solution $x$. Let $y\in\mathbb F^m\setminus0$. Then $x\otimes y$ and $y\otimes x$ are nonzero but $(A\otimes B)(x\otimes y)=(Ax)\otimes(By)=0$ and likewise $(B\otimes A)(y\otimes x)=0$. Therefore $A\otimes B$ and $B\otimes A$ are singular and their determinants are zero.