Rank normal form of a matrix

There is a standard result in matrix theory that goes like this: Suppose $A$ is an $m\times n$ matrix of rank $r$, then there exist two non-singular matrices $E$ (of size $m\times m$) and $F$ (of size $n\times n$) such that $A=ENF$ where $N=\pmatrix{I_r&0\\ 0&0}$ that is $I_r$ is the identity matrix of order $r$ and the $0$s here are null matrices. But how to prove this? Please help.

Solution 1:

Hint:

Row operations on the matrix $A$ can be expressed by nonsingular square matrices on the left of $A$, and similarly column operations can be expressed as nonsingular square matrices on the right of $A$. If this is unfamiliar to you, you should take a look at the wiki article on the subject.

Using row and column operations, you can reduce $A$ to the form of $N$ like this: $XAY=N$. To get $E$ and $F$ in your description above, you just need to invert $X$ and $Y$ to get: $A=X^{-1}NY^{-1}$. Those are your choices for $E$ and $F$.