Invertible matrix implies square matrix

Question

Given an $m\times n$ matrix $A$ and an $n\times m$ matrix $B$ such that $AB=I_m$ and $BA=I_n$, I would like to show that, in fact, $m=n$.

(The $m$ and $n$ may be the other way around. They should be such that $A$ and $B$ are inverses, as per the title of the question)

I know this can be done easily by seeing $A$ and $B$ as linear transformations between vector spaces, but I am wondering if anyone is aware of a proof that only involves matrix algebra. That is, a proof that would work even if the entries of $A$ and $B$ were merely from a unital ring rather than a field (although I am not even sure that the statement would be true in that case, but I haven't been able to find a counterexample).

Answer 1

Here is another way: if $A$ and $B$ are any two matrices that can be multiplied from both sides then $tr(AB)=tr(BA)$. You can easily prove it using the definition of matrix multiplication. Hence if $AB=I_m$ and $BA=I_n$ then $m=tr(I_m)=tr(AB)=tr(BA)=tr(I_n)=n$.

Answer 2

Here is one way: denote the row space of matrix $A$ by $Row(A)$ and the column space by $Col(A)$. Firstly, by the properties of the matrix product, show that $Col(AB)\subseteq Col(A)$ and $Row(AB)\subseteq Row(B)$.

Now observe that if $AB=I_m$ then $F^m\subseteq Col(A)$, but as $A$ has $n$ columns, it follows that $n\leq m$. Similarly, if $BA=I_n$ then $F^n\subseteq Row(A)$, but as $A$ has $m$ rows it followas that $m\leq n$. Hence $m=n$.

Answer 3

Edit, answering the edited question.

Assume that $m<n$, wlog.

Then $\operatorname{rk}A\leq m$ and therefore $\operatorname{rk}AB\leq m$ and $\operatorname{rk}BA\leq m$. It follows that none of these products can be $I_n$.