Non square matrix problem Do there exist  $2$ non square matrices $A$ and $B$ such that both products $AB$ and $BA$ are defined and are identity matrices (of course of different orders)? 
 A: No, there do not exist two such non-square matrices. Let $A$ be a matrix with $m$ columns and $n$ rows ($m>n$) and $B$ with $n$ columns and $m$ rows. Add zero row or column vectors to $A$ and $B$ so as to make them square matrices. Then $AB=\begin{pmatrix}I_n&0\\ 0& 0\end{pmatrix}$ and $BA=I_m$ have different traces, a contradiction.
A: The answer is no, if EDIT $m,n >1$:
Let $A$ wolg be $ n \times m$ and let $B$ be $m\times n$ ( needed for product matrix to be square) ; $n \neq m $ with EDIT $ n,m \geq 2$. Then $B$ has a non-trivial kernel, meaning there is a non-zero vector $v$ with $Bv=0$. Then $AB$ is an $m \times m$ matrix with $ABv=A(Bv)=A.0=0$ for a nonzero vector $v$, so that $AB$ cannot be the identity. A similar argument shows that $BA$ cannot be the identity matrix either.
EDIT: The  $ (N+1) \times n$  matrix $B= [I_n 0 ]^T $ is not a counterexample:
Given any vector $v= [0 0 ..... x ]^T$ with n 0 entries and the $(n+1)$st entry any number , then $$ Bv= [ I_n 0 ] ^T [ 0 0...x] =0 $$
EDIT 2: My point did not go through for some reason . It is not possiblefor both $AB$ and $BA$ be the identity: $A$ is $m \times n$, $B$ I $n \times m $ , then by basic arithmetic, either $m>n$ or $n>m$ , say $n>m$ wolg. Then $B$ has non-trivial kernel $K$. Say $v \neq 0, Bv=0$ . Then $ABv= A(Bv)= A0=0 $ , but $Id.v =v \neq 0 $, so AB cannot be the identity.
