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It is well known that if $A,B$ are square matrices of size $n$ (with terms in some field $\mathbb{K}$) such that $AB=I_n$ (the unit matrix of size $n$) then $BA=I_n$.

This can be proved by considering the endomorphisms $f,g$ of $\mathbb{K}^n$ which are represented respectively by $A,B$ in the canonical basis of $\mathbb{K}^n$ :

Since $f\circ g=id$, we can see that $f$ is surjective and $g$ is injective, hence both are isomorphisms. Now we have :

$g\circ f=g\circ f\circ (g\circ g^{-1})=g\circ (f\circ g)\circ g^{-1}=g\circ g^{-1}=id$

which proves that $BA=I_n$.

Of course, it can also be seen as follows : $\det(A)\,\det(B)=\det(AB)=1$ hence $\det(A)$ and $\det(B)$ are nonzero.

Now my question ...

How to prove the same result, using only algebraic calculations (no determinant, no rank formula) ?

Adren
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  • You cannot. See https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – user1551 Nov 18 '20 at 17:49
  • Everything should be made as simple as possible, but not simpler – Dunham Nov 18 '20 at 18:33
  • Your requirement basically forces one to examine the matrix multiplication formula. I remember asking this exact question back in high school to myself and concluded that you cannot. It was back at a time when notes were not electronic, unfortunately. – Argyll Nov 18 '20 at 22:15
  • If there was a purely algebraic proof, then it would be true for all algebras. But the assertion is false for infinite dimensional spaces. – Chrystomath Nov 20 '20 at 05:02
  • @Chrystomath : Obviously, and I know such counter-examples, but one can maybe imagine some calculation involving explicitly the terms of those matrices (and thus using explicitly the fact that the algebra is finite dimensional). – Adren Nov 20 '20 at 05:43

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