Here is an observation I made:
If A and B are two non square matrices conformable for multiplication then either $\det(AB) =0$ or $\det(BA)=0$ or both.
I could come up with this proof:
Let $A_{m\times n}$ and $B_{n \times m}$. WLOG, let $m < n$. Then $\text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B)) \leq \min(m,n) = m$. Simillarly, $\text{rank}(BA) \leq \min(\text{rank}(A), \text{rank}(B)) \leq \min(m,n) = m$. But order of $AB=m$ and order of $BA = n$. Hence matrix $BA$ has rank less than its order.
Is this proof(and consequently the theorem) correct? And are there any alternative proofs?
Best Answer
Yes. Basically if $A$ is $m\times n$ and $B$ is $n\times m$, $m<n$ then $A$ can be seen a a linear function from $n$ dimensions into $m$ dimensions and $B$ can be seen as a function of $m$ dimensions into $n$ dimensions.
Thus the image of $B$ is a space of degree $n$ and of dimension at most $n$. Thus the $A(\mathrm {Im} B)$ is a space of degree $m$ but dimension at most $n$.