If $A,B\in \mathbb R^{n\times n}$ s.t. $AB=I$ then $BA=I$.

Asked Jul 26 '18 at 20:47

Active Jul 26 '18 at 21:34

Viewed 33 times

I want to prove that if $A,B\in \mathbb R^{n\times n}$ are s.t. $AB=I$ then also $BA=I$ (where $I$ is the identity matrix).

$$AB=I\implies BABA=BA\implies (BA)^2-BA=0\implies BA(BA-I)=0.$$

How can I conclude from here ? I know that $CD=0\not\Rightarrow C=0$ or $D=0$. So I can't continue. I also thought to do : $$BA(BA-I)=0\implies ABA(BA-I)=0\underset{AB=I}{\implies} A(BA-I)=0,$$ but it's not conclusive. Any idea ?

Edited

I can't use determinant.

edited Jul 26 '18 at 21:34

asked Jul 26 '18 at 20:47

Peter

1,005

@GonzaloBenavides: What do you mean ? I have a theorem that say $A$ invertible at left $\iff$ $A$ invertible at right $\iff$ $A$ invertible both side. – Peter Jul 26 '18 at 20:59
@Peter Yes, but nowhere in the proposition is it assumed that $A$ and $B$ are non-singular (invertible). – Sean Roberson Jul 26 '18 at 21:01
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@SeanRoberson: Do you know singular matrix s.t. $AB=I$ ? – Peter Jul 26 '18 at 21:04
Since $AB=I$, then $\det A \det B = 1$ and therefore $A$ is not singular. – Gonzalo Benavides Jul 26 '18 at 21:11
Try multiplying by $A^{-1}$ both sides of the last equation. – Gonzalo Benavides Jul 26 '18 at 21:12
@GonzaloBenavides: I can't use det(A). I have to prove it algebraically. Moreover, $A^{-1}$ a priori not exist. I need an algebraic proof. – Peter Jul 26 '18 at 21:13
You should tell us what we can use. – Gonzalo Benavides Jul 26 '18 at 21:15

If $A,B\in \mathbb R^{n\times n}$ s.t. $AB=I$ then $BA=I$.

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