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.

  • 1
    $\begingroup$ If $A$ and $B$ are matrices, then each is a matrix, not matrice. $\endgroup$ – Omnomnomnom Apr 4 '17 at 19:34
  • 3
    $\begingroup$ 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

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 $B$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.