Why can this product of rectangular matrices not be the identity matrix?

Question

Let $m > n$ be positive integers. Show that there do not exist matrices $A ∈ R_{m×n}$ and $B ∈ R_{n×m}$ such that $AB = I_m$, where $I_m$ is the $m × m$ identity matrix

If someone could explain it with an example, it will be appreciable.

Answer 1

It's a classic result that if $f\circ g$ is a bijective map then $f$ is surjective and $g$ is an injective map, hence using this and assuming the existence of $B$ we see that the linear transformation $$f\colon\Bbb R^n\rightarrow \Bbb R^m,\quad x\mapsto Ax$$ is surjective i.e $\operatorname{Im}g=\Bbb R^m$. Moreover by the rank-nullity theorem we have $$\dim\Bbb R^n=n\ge\dim\operatorname{Im}g=m$$ which contradicts the hypothesis $n<m$. Conclude.

Answer 2

Each column of $AB$ is a linear combination of the rows of $A$, and each row of $AB$ is a linear combination of the columns of $B$. So $\operatorname{rank}(AB) = \dim(\operatorname{colspace}(AB))\leqslant\operatorname{rank}(A)$, and $\operatorname{rank}(AB) = \dim(\operatorname{rowspace}(AB))\leqslant\operatorname{rank}(B)$. So $\operatorname{rank}(AB)\leqslant\min\{\operatorname{rank}(A),\operatorname{rank}(B))\leqslant\min\{m,n\} = n < m$, but $\operatorname{rank}(AB) = \operatorname{rank}(I_m) = m$. Contradiction.