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I don't recall this well from linear algebra so I just need some quick confirmation.

If $A$ is an $n \times n$ matrix and B is also $n \times n$ and $AB=I_n$ does it follow that also $BA=I_n$?

I think no but then what do we call inverse matrix: the left inverse, or the right one, or only when the left and right are the same we talk about inverse matrix of $A$?

Also, do we define inverse matrices of non-square matrices?

Rodrigo de Azevedo
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peter.petrov
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  • Yes, $AB=I$ implies $BA=I$. (when $A$ and $B$ are both $n\times n$ matrices). So if a matrix is invertible from one side then it is invertible. For non square matrix we usually do not define an inverse. – Mark Aug 04 '20 at 09:21
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    Does this answer your question? [If $AB = I$ then $BA = I$](https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i) – Aman Pandey Aug 04 '20 at 09:23
  • @AmanPandey Seems it does but the proof doesn't seem very simple. Isn't there some simpler proof? – peter.petrov Aug 04 '20 at 09:25

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Using the fact that $\det{(AB)}=\det{(A)}\det{(B)}$ we can conclude that both $A$ and $B$ have non-zero determinants and hence are invertible. So we can write \begin{align} AB=I_n &\iff A^{-1}AB=A^{-1}I_n\\ &\iff B=A^{-1}\\ &\iff BA=A^{-1}A\\ &\iff BA=I_n\\ \end{align} So that a left inverse of any square matrix is also a right inverse and vice versa.

Peter Foreman
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  • @peter.petrov [Here](https://math.stackexchange.com/questions/60284/how-to-show-that-detab-deta-detb/60357) are some proofs of this fact. – Peter Foreman Aug 04 '20 at 09:30