Prove that if $A$ is an m x n matrix, there is an invertible matrix $C$ such that $CA$ is in reduced-row echelon form Does anyone have any hints or answer how to go about this?
$C$ must be an m x m matrix. Thus the resulting reduced-row echelon form $CA$ is m x n.  
I originally had $CA = I_n$, but then I realised that reduced-row isn't necessarily identity matrix and in this case, cannot be because of the m x n size.
I'm stuck.
 A: When you perform Gauss-Jordan elimination to take a matrix $A$ into reduced row-echelon form, you perform one of the following elementary row operations on $A$:


*

*Scale the $i$-th row of $A$ by a nonzero scalar $c$

*Switch row $i$ of $A$ and row $j$ of $A$

*Add $c \cdot (\text{row $j$})$ to row $i$


Each of these row operations can be represented by an elementary matrix $E$:


*

*A diagonal matrix with the $i$-th term equal to $c$ and all others to $1$

*The identity matrix with row $i$ and row $j$ switched

*The identity matrix with entry $(i,j)$ set to $c$. 


Since the row operations are invertible, the elementary matrices are invertible, and their inverses are also elementary matrices, corresponding to the inverse elementary operation.
As you perform a step of Gauss-Jordan elimination, the state of the matrix after performing an elementary row operation with elementary matrix $E$ to $A$ is $EA$, so our reduced-row echelon form can be written as:
$$ R = E_k\cdots E_3E_2E_1 A$$
where $E_i$ is the elementary matrix corresponding to the $i$-th elementary row operation taken in Gauss-Jordan elimination. Therefore, the invertible matrix you are looking for is:
$$ C = E_k \cdots E_3 E_2 E_1 $$
This matrix is invertible because it is the product of invertible matrices.
