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I am aware that invertible square matrices are left invertible and right invertible, and that the left and right inverses are equal. However, I was wondering whether exists a non square $m\times n$ matrice $A$, so that exist both:

  1. An $n\times m$ matrice $B$ so that $AB = I_m$
  2. An $n\times m$ matrice $C$ so that $CA = I_n$

I just couldn't think of an example nor of a proof that these two conditions provide that A is necessarily square.

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    $\begingroup$ If $A$ and $B$ are matrices, then each is a matrix, not matrice. $\endgroup$ – Omnomnomnom Apr 4 '17 at 19:34
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    $\begingroup$ Note that $\operatorname{rank}(AB) \leq \operatorname{rank}(A) \leq \min\{m,n\}$ and $\operatorname{rank}(CA) \leq \operatorname{rank}(A) \leq \min\{m,n\}$. $AB$ and $CA$ cannot both have full rank. $\endgroup$ – Omnomnomnom Apr 4 '17 at 19:36
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Suppose you have an $n\times m$ matrix $A$ with $n\neq m$.

If $n<m$ then $rank(BA)\leq rank(A)\leq n$ and so $BA\neq I_m$ for all $B$.

If $m<n$ then $rank(AB)\leq rank(A)\leq m$ and so $AB\neq I_n$ for all $B$.

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  • $\begingroup$ Awesome, thanks. $\endgroup$ – Noa Apr 6 '17 at 13:15

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