I am interested to write a rigorous proof for a well-known results - every elementary operation is invertible. Although, there is another SE post that proves this result, I am not familiar with abstract algebra, so I found the approach difficult to follow. And I want to avoid any circularity(using an advanced results for example, to prove a more elementary result).
It would be nice, if someone could give any tips/hints on how to complete the proof or if my proof is correct.
Every elementary matrix $E$ is invertible.
Proof. (My attempt).
Let $E$ be an elementary matrix of order $n$ obtained by performing elementary column operations on $I_n$. Intuitively, any column of $E$ is a unique linear combination of the columns of $I_n$.
Let $E_j$ be the $j$th column of $E$.
Suppose $E_j= \begin{bmatrix} a_{1j}\\ a_{2j}\\ a_{3j}\\ \vdots \\ a_{nj} \end{bmatrix}=a_{1j}e_1 + a_{2j}e_2 + a_{3j}e_3+\ldots + a_{nj}e_n$.
$E_j$ has a unique representation in $\mathbb{R}^n$. For if we assume,
$E_j=b_{1j}e_1 + b_{2j}e_2 + b_{3j}e_3+\ldots + b_{nj}e_n$
Then:
$(a_{1j}-b_{1j})e_1+(a_{2j}-b_{2j})e_2+\ldots + (a_{nj}-b_{nj})e_n=0$
As $\{e_1,e_2,\ldots,e_n\}$ are linearly independent, we obtain $a_{1j}=b_{1j}, a_{2j}=b_{2j},\ldots,a_{nj}=b_{nj}$.
I am not sure how to proceed from here.
I could say, because each $E_j$ has unique coordinates, each $E_j \notin \text{span}(E_1,E_2,\ldots,E_{j-1})$. Therefore, they form a linearly independent set.
Invoking the inverse matrix theorem, $E$ must be invertible.