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I was trying to come up with a proof of why: $AA^{-1} = I$.

If we know that: $A^{-1}A = I$, then $A(A^{-1}A) = A \implies (AA^{-1})A = A$.

However I don't like setting $AA^{-1} = I$ for fear that it might be something else at this point, even though we know that $IA=A$. For example, could $A$ times its inverse equal something other than the identity leading back to the original matrix $A$.

Does anyone have a another proof for why $A$ times its inverse would give you the identity or could explain something I'm missing?


marked as duplicate by almagest, hardmath, user223391, Oscar Cunningham, Crostul Jun 16 '16 at 14:15

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    $\begingroup$ How do you define $A^{-1}$ if not that it is $AA^{-1}=I$? $\endgroup$ – Hirshy Jun 16 '16 at 13:54
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    $\begingroup$ You have to use that these are effective matrices: associativity and existence of the identity are not enough to prove that right invertibility is equivalent to left invertibility. $\endgroup$ – Crostul Jun 16 '16 at 13:55
  • $\begingroup$ How do you know that these are effective matrices then? $\endgroup$ – Will Jun 16 '16 at 13:56

We say a matrix $B$ is an inverse for $A$ if $AB = BA = I$, and the notation for $B$ is $A^{-1}$.
So it's by definition $AA^{-1}=I$, you cannot really prove it.

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    $\begingroup$ The question is asking why $AB=I$ implies $BA=I$. The use of the word "inverse" or the symbol $A^{-1}$ is a red herring. $\endgroup$ – almagest Jun 16 '16 at 14:15

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