Two real rectangular matrices $AB=I$ Let $A$ be an $m×n$ real valued matrix and $B$ be an $n×m$ real valued matrix so that $AB$=$I$. Thus we must have


*

*$n>m$

*$m\ge n$

*if $BA=I$ then $m>n$

*either $BA=I$ or $n>m$
What I tried: Either $n<m$ or $n>m$ → either $\operatorname{Rank}(A) = \operatorname{Rank}(B)=n$ or $n>m$ → either $BA=I_n$ or $n>m$. So option 4 seems to be true to me. But how can I reject option 3?

Answering my own question for the sake of completeness for the learners in mathematics community, the proof is as follows:
$Rank(AB)=Rank(I_m)=m\implies m\le n$
$\therefore$ either $m<n\ or\ m=n$
In case $m=n$;$AB=I\implies B=A^{-1}\implies BA=I$
Thus, either $m<n\ or\ m=n$ is true.
