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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

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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 $E_n E_{n-1} \ldots E_1 A = I$.

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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).

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  • $\begingroup$ Wait, why does the system being equivalent to Ix= 0 mean exactly that [A0] can be transformed into [I0]? Maybe I'm not clear on exactly what adjoining two matrices means...I thought you could just arbitrarily put any two matrices next to each other and that meant you had adjoined them. I didn't realize it implied any kind of equality such as being equivalent to Ix = 0...? $\endgroup$ – dagny Aug 15 '16 at 22:02

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