TITLE: Proof for '$AB = I$ then $BA = I$' without Motivation?
QUESTION [12 upvotes]: I have read this question page (If $AB = I$ then $BA = I$) by Dilawar and saw that most of proofs are using the fact that the algebra of matrices and linear operators are isomorphic.
But from a simple view, matrix is only an structured set of scalars, and the fact that the dot product of the i-th row of $A$ and the j-th column of $B$ equals to the Kronecker delta is just a componentwise algebraic informarion.
Then I started to wonder if there's any "brutal" proof that does not visit the "higher" domain of algebraic structures and just uses the simple componentwise algebraic operations to prove that  the dot product of the i-th row of $B$ and the j-th column of $A$ equals to the Kronecker delta from the given condition. (Proof that even a "machine" can do)
Should we think a matrix as more than a mere 'number box' to show $BA=I$?

REPLY [5 votes]: Here is a sketch of a "brutal proof" of the sort you imagine. (On rereading the other answers, you've already seen this. But the notes at the end may be interesting.) A more detailed version of the following can be found in "Bijective Matrix Algebra", by Loehr and Mendes. 
Remember that the adjugate matrix, $\mathrm{Ad}(A)$, is defined to have $(i,j)$ entry equal to $(-1)^{i+j}$ times the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting the $i$-th row and $j$-th column from $A$.
Write down a brute force proof of the identity:
$$\mathrm{Ad}(A) \cdot A = A \cdot \mathrm{Ad}(A) = \det A \cdot \mathrm{Id}_n$$
by grouping like terms on both sides and rearranging.
Likewise, write down a brute force proof of the identity
$$\det(AB) = \det(A) \det(B).$$
So, if $AB=\mathrm{Id}$, you know that $\det(A) \det(B)=1$.
Now compute $\mathrm{Ad}(A) ABA$ in two ways:
$$(\mathrm{Ad}(A) A)BA = \det(A) BA$$
and
$$\mathrm{Ad}(A) (AB) A = \mathrm{Ad}(A) A = \det(A) \mathrm{Id}.$$
Since $\det(A) \det(B) =1$, you also have $\det(B) \det(A)=1$, and get to deduce that $BA = \mathrm{Id}$.
There is some interesting math here. Let $R$ be the polynomial ring in $2n^2$ variables $x_{ij}$ and $y_{ij}$ and let $X$ and $Y$ be the $n \times n$ matrices with these entries. Let $C_{ij}$ be the entries of the matrix $XY-\mathrm{Id}$ and let $D_{ij}$ be the entires of the matrix $YX-\mathrm{Id}$. Tracing through the above proof (if your subproofs are brutal enough) should give you identities of the form $D_{ij} = \sum_{k,\ell} P_{k \ell}^{ij} C_{ij}$. It's an interesting question how simple, either in terms of degree or circuit length, the polynomials $P_{k \ell}^{ij}$, can be made. 
I blogged about this question and learned about some relevant papers (1 2), which I must admit I don't fully understand.