Prove that $A$ and $B$ are invertible and $B=A^{-1}$ Suppose $A$ and $B$ are $n \times n$ matrices.  Assume $AB=I$. Prove that $A$ and $B$ are invertible and that $B=A^{-1}$.
Please let me know whether my proof is correct and if there are any improvements to be made.
Assume $AB=I$.  Then $(AB)A=IA=A$. So, $A(BA)=AI=A$.  Then $BA=I$.  Therefore $AB=BA=I$.  Thus  $A$ and $B$ are invertible.  And by definition $B=A^{-1}$, so $AB=AA^{-1}=I$.
It doesn't seem quite correct.  Thanks in advance for any advice.
 A: Proof #1: (along the lines mentioned in the comments)
As $AB=I$, you know that $A$ is onto as a linear transformation, because $x=Ix=ABx=A(Bx)$ for any $x\in\mathbb{R}^n$. This implies that $A$ is bijective, being a surjective linear transformation in a finite-dimensional space. So there exists $A^{-1}$. Now
$$
A^{-1}=A^{-1}I=A^{-1}AB=B.
$$
Proof #2 (using determinants)
Since $AB=I$, we have
$$
1=\det I=\det AB=\det A\,\det B.
$$
So $\det A\ne0$ and $A$ is invertible, and again we can do 
$$
B=IB=A^{-1}AB=A^{-1}I=A^{-1}.
$$
A: A good first step would be to look at some of the answers to this question. The accepted one, by Davidac897, is pretty elementary and is probably the place to start. You’re almost certainly not yet ready for Martin Brandenberg’s answer, and I’d also skip Bill Dubuque’s answers for now: they’re also aimed at someone with more background. The proof given by falagar, on the other hand, is well worth a look, and you should certainly look at Blue’s answer, which is deliberately very elementary.
A: Suppose $AB = I$.  ($A$ and $B$ are $n \times n$ matrices.)
First note that $R(A) = \mathbb{R}^n$.  (If $y \in \mathbb{R}^n$, then
$y = Ax$, where $x = By$.)
It follows that $N(A) = \{0\}$.
We wish to show that $BAx = x$ for all $x \in \mathbb{R}^n$.
So let $x \in \mathbb{R}^n$, and let $z = BAx$.
Then $Az = A(BAx) = (AB)Ax = Ax$, which implies that
$z = x$ because $N(A) = \{ 0 \}$.  So $BAx = x$.
