The ring of matrix is not an integral domain. How to prove that the inverse is unique?

  • $\begingroup$ You don't need it to be an integral domain to have unique inverses. $\endgroup$ – mez May 22 '14 at 18:34

If $ab=ba=1$, and if also $ac=ca=1$, you have $c=(ba)c=b(ac)=b$. In fact, if you reread it, this shows that whenever you know both a left inverse exists and a right inverse exists, then actually they are the same element, so it is a two-sided inverse and it is unique.

It has little to do with integral domains: it's true for any associative operation with an identity.

  • 1
    $\begingroup$ You could also mention that your argument actually shows that if a left inverse and a right inverse exist, then they are equal and so he inverse exists and is unique. $\endgroup$ – Martin Argerami May 22 '14 at 18:52
  • $\begingroup$ @MartinArgerami Good idea: I added that. Thanks! $\endgroup$ – rschwieb May 22 '14 at 19:17

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