If $AB = I$ then $BA = I$ If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$. 


I do not understand anything more than the following.


Elementary row operations.
Linear dependence.
Row reduced forms and their relations with the original matrix.


If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?

P.S.: Please avoid using the transpose and/or inverse of a matrix.
Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence.

Suppose each matrix is $n$ by $n$. We consider our matrices to all be acting on some $n$-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and $n$ by $n$ matrices).

Then $AB$ has range equal to the full space, since $AB=I$. Thus the range of $B$ must also have dimension $n$. For if it did not, then a set of $n-1$ vectors would span the range of $B$, so the range of $AB$, which is the image under $A$ of the range of $B$, would also be spanned by a set of $n-1$ vectors, hence would have dimension less than $n$.

Now note that $B=BI=B(AB)=(BA)B$. By the distributive law, $(I-BA)B=0$. Thus, since $B$ has full range, the matrix $I-BA$ gives $0$ on all vectors. But this means that it must be the $0$ matrix, so $I=BA$.
The OP from 10 years ago said he understood
1.Elementary row operations.
2.Linear dependence.
3.Row reduced forms and their relations with the original matrix.

I'll use only a bit more than (1) and (3). Each elementary row operation corresponds to left multiplication by an elementary matrix and every elementary matrix is invertible. I won't use (2) but I will use
(4) If the rref of square matrix has a zero row, then it does not have a left inverse.

Proof. Let square $H$ have rref $R$ whose last row is zero. Setting a parameter equal to 1 and using $R,$ there is a nonzero column vector $X$ such that $HX={\bf 0}.$ Suppose $KH=I.$ Then
$$ X=IX=KHX=K(HX)=K{\bf{0}}={\bf 0},\ \text{a contradiction.}$$
Let $AB=I.$ By (4), $I$ is the rref of $B.$  Therefore, $CB=I,$ where $C$ is a product of elementary matrices. By taking, in reverse order, the product of the inverses of these elementary matrices, it follows that $C$ is invertible. $$C^{-1}=C^{-1}(CB)=IB=B$$
so that $BC=I.\tag 1$   $$A=AC^{-1}C=(AB)C=IC=C\tag 2 $$
Combining (1) and (2), we conclude that $BA=I.$
(I just noticed the P.S. of the OP. Sorry. Also, there's probably overlap with earlier proofs. I just wanted to give a short, simple proof that did not use any concepts from vector spaces.)