Can someone explain this part of this property of invertible matrices proof? The property states, "A square matrix A is invertible iff it can be written as the product of elementary matrices"
I'm confused on the part of the theorem where they're trying to show that if A is invertible, then it can be written as the product of elementary matrices.
This is that section of the proof:
"Assume A is invertible. You know the system of linear equations represented by Ax=0 has only the trivial solution. But this implies that the augmented matrix [A 0] can be rewritten in the form [I 0] (using elementary row operations corresponding to E1,E2,...,Ek). So, Ek,...,E2,E1A I and it follows that A = E1-1E2-1...Ek-1 . A can be written as the product of elementary matrices."
I just don't get how knowing that Ax=0 has only the trivial solution implies that [A 0] can be written in the form [I 0]. Wasn't it already obvious that A can be rewritten as I since it's invertible? And obviously if there's a 0 matrix adjoined A to it it's going to stay the zero matrix no matter what row operations are done on it? What's the point of doing that?
I'm just generally confused on this proof
 A: The point is that each step in the process of Gauss-Jordan elimination corresponds to multiplying your matrix on the left by an elementary matrix.
If you start with $[A\mid b]$ (where $A$ is your matrix and $b$ the augmented column), you get $[E_1 A \mid E_1 b$ in the first step, for some elementary matrix $E_1$, then $[E_2 E_1 A \mid E_2 E_1 b]$ for some elementary matrix $E_2$, and so on.
The "augmented" column is not important, the non-augmented part is.  If the matrix is invertible, at the end of Gauss-Jordan elimination you get 
$[ I \mid something]$.  That is, Gauss-Jordan elimination ends by telling you a unique value for each variable.
And this says that $E_n E_{n-1} \ldots E_1 A = I$.
A: If the system $Ax=0$ hasn't a unique solution, then it has infinitely many (it cannot have no solution, because the zero vector is always a solution). In this case, row operations end by giving a matrix having its last line(s) entirely filled of $0$ and you can't obtain $[I\; 0]$. 
Now, if $Ax=0$ has a unique solution, then this solution is necessarily the zero vector. Hence, the system is equivalent to $x=0$ (or $Ix=0$). This means exactly that $[A\; 0]$ can be transformed (via elementary row operations) into $[I\; 0]$. 

Wasn't it already obvious that A can be rewritten as I since it's invertible? 

It was obvious that there is a matrix $P$ s.t. $PA=I$ but it wasn't obvious at all that there is some $P$ being a product of elementary row matrices s.t. $PA=I$ (this could have existed for only certain particular kind of invertible matrices). 
