Assuming $AB=I$ prove $BA=I$ Possible Duplicate:
  If $AB = I$ then $BA = I$  




Most introductory linear algebra texts define the inverse of a square matrix $A$ as such:

Inverse of $A$, if it exists, is a matrix $B$ such that $AB=BA=I$.

That definition, in my opinion, is problematic. A few books (in my sample less than 20%) give a different definition:

Inverse of $A$, if it exists, is a matrix $B$ such that $AB=I$. Then they go and prove that $BA=I$. 

Do you know of a proof other than defining inverse through determinants or through using rref?  

Is there a general setting in algebra under which $ab=e$  leads to $ba=e$ where $e$ is the identity?
Multiply both sides of $AB-I=0$ on the left by $B$ to get
$$
(BA-I)B=0\tag{1}
$$
Let $\{e_j\}$ be the standard basis for $\mathbb{R}^n$. Note that $\{Be_j\}$ are linearly independent: suppose that
$$
\sum_{j=1}^n a_jBe_j=0\tag{2}
$$
then, multiplying $(2)$ on the left by $A$ gives
$$
\sum_{j=1}^n a_je_j=0\tag{3}
$$
which implies that $a_j=0$ since $\{e_j\}$ is a basis. Thus, $\{Be_j\}$ is also a basis for $\mathbb{R}^n$.

Multiplying $(1)$ on the right by $e_j$ yields
$$
(BA-I)Be_j=0\tag{4}
$$
for each basis vector $Be_j$. Therefore, $BA=I$.

Failure in an Infinite Dimension

Let $A$ and $B$ be operators on infinite sequences. $B$ shifts the sequence right by one, filling in the first element with $0$. $A$ shifts the sequence left, dropping the first element.

$AB=I$, but $BA$ sets the first element to $0$.

Arguments that assume $A^{-1}$ or $B^{-1}$ exist and make no reference to the finite dimensionality of the vector space, usually fail to this counterexample.
For $A$ and $B$ square, $AB=I$ implies $(AB)^{-1} = B^{-1}A^{-1}=I$. Then multiply $B$ from the left side, $A$ from the right side. $BB^{-1}A^{-1}A=BIA$ implies $BA=I$.