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Let $A\in\mathbb R^{n\times n}$.

1) If it's invertible at left then it's invertible.

2) If it's invertible at right, then it's invertible.

I saw several proofs (here and here), but they look much more complicated than what I did. So I'm I correct and if no, why ?

For 1) (because 2) is the same). Way 1 : We have that $\mathbb R^n\to \mathbb R^n, v\longmapsto Av$ is injective. By rank theorem it's also surjective. Therefore, there is $C\in\mathbb R^{n\times n}$ s.t. $AC=I$. But $$B=B(AC)=(BA)C=C,$$ and thus $A$ is invertible.

Way 2 : Let $B\in \mathbb R^{n\times n}$ s.t. $BA=I$. In particular, $\mathbb R^n\to \mathbb R^n; v\longmapsto Bv$ is surjective. By rank theorem it's also injective, and thus $$B(AB-I)=(BA)B-B=B-B=0\implies AB-I=0,$$ and thus $AB=I$, i.e. $A$ is invertible.

Way 3 :

Without using rank theorem. We have that $$BA=I.$$ We have $$B(AB-I)=0.$$ Since $B$ is invertible at right, it's surjective and thus full rank. Therefore, there are $P,Q\in\mathbb R^{n\times n}$ invertible s.t. $PBQ^{-1}=I$, i.e. $B=P^{-1}Q$. Therefore $$B(AB-I)=P^{-1}Q(AB-I)=0\implies AB-I=0\implies AB=I.$$

Does it work ?

Henri
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1 Answers1

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Yes, your proof is correct. The equivalence between invertible at left and invertible at right is a simple consequence of rank theorem.

Pjonin
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