Given two square matrix matrices ($A$ and $B$ both have the same order) such that $AB = I$ is it true that $BA=I$?

Question

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

Answer 1

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 $A = B^{-1}$.

Answer 2

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 $BA = I$.