Say $A$ and $B$ are square matrices with $K$-valued entries, $K$ a field. There is an 'abstract' argument showing this implies $BA=I$, namely that $AB=I$ implies $A$ is surjective, so it is injective by rank-nullity, so its right inverse must be its left inverse. My question is whether there is some other, more 'numerical' argument showing this, since obviously one can interpret $AB=I$ as some system of equations. The argument should use (I think) the fact that $K$ is a field, since rank-nullity does not work over general modules. So somewhere in the argument division should be involved. Thanks in advance.
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You are basically asking for Cramer's rule. – MooS Aug 01 '17 at 09:59
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1You can find many, many answers in this thread. – Viktor Vaughn Aug 01 '17 at 10:24
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2Possible duplicate of If $AB = I$ then $BA = I$ – Arctic Char Dec 03 '19 at 02:24
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An algebraic argument in terms of equations ("numerical") is possible, using the determinant. Since $1=\det I=\det(AB)=\det(A)\det(B)$, we know that $\det(A)\neq 0$ so that the inverse $A^{-1}$ exists. Hence $AB=I$ gives $B=A^{-1}AB=A^{-1}$, so that $BA=A^{-1}A=I$.
Dietrich Burde
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This is nice. I wonder whether there is an even more elementary approach. – M. Van Aug 01 '17 at 10:03
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@M.Van No; I was suggesting a simplification of the part of the argument that comes after invertibility is established. – J.G. Aug 01 '17 at 15:50