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my try:

$AB = I$

$BAB = B$

$BAB = B$

$BAB - B= 0$

$(BA - I)B= 0$

I wanted to be able to multply by A and appear AB somewhere up there...

$AB = I$

$ABA = A$

$A(BA - I) = 0$

mmm.... can someone gimme a hint at least ? I know we can't say that because $B$ is not null (although this is true) then $BA-I$ must be null

onlyme
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    You won't be able to prove it using algebraic manipulations alone; it's not true for more general rings. – Dustan Levenstein Mar 23 '15 at 15:50
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    A purely algebraic proof of this is kind of tedious. I think the easiest way to do it is to say that if $AB=I$ then $A$ represents a surjective linear map $\mathbb{R}^n \to \mathbb{R}^n$ which is thus injective. Then the only candidate for a left inverse is the right inverse, i.e, $B$. – shalop Mar 23 '15 at 15:52

2 Answers2

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Suppose that $B\vec x = \vec 0$. Then $AB \vec x = \vec 0$ so that $\vec x = \vec 0$ since $AB = I$. This means that $B\vec x = \vec 0$ has only the trivial solution $\vec x = \vec 0$, and so must be invertible. Then you have $$A = AI = ABB^{-1} = IB^{-1} = B^{-1}$$ so that $A = B^{-1}$.

Umberto P.
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Let $n$ be the rank. Then a square matrix $M$ induces a linear map $T_M : \mathbb R^n \rightarrow \mathbb R^n $. Note $T_A \circ T_B = T_{AB} = Id_{\mathbb R^n}$.

Hence $T_A = (T_B)^{-1}$ and $ T_{BA} = T_B \circ T_A = Id_{\mathbb R^n} $. Therefore we have $BA = I$.

Basics
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