We shall show that
every matrix
has unique RREF. More precisely, we shall show the following.
If $B$ and $C$ have the same size, are both in RREF and
have the same row space, then $B=C.$
Our argument is a slightly simplified version of a proof by Thomas Yuster.
Let $B$ and $C$ be $m\times n.$ Since they have the same row space, so they must
have the same rank, $r,$ say. Hence they must both have the
first $r$ rows nonzero, and the last $m-r$ rows null.
We shall apply induction on $n.$ Trivial for $n=1.$
By induction hypothesis, $B$ and $C$ must have the
first $n-1$ columns identical.
Shall show that the $n$-th columns of $B$ and $C$
must also match. This will complete the proof.
For this it is enough to show that if $b_{in}\neq 0$ then $c_{in}=b_{in}.$
Let $b_{in}\neq 0.$ Then $i\in\{1,...,r\}.$
The $i$-th row of $B$ must be a linear combination of
the top $r$ rows
of $C.$ But due to RREF structure, these rows of $C$
(except the $i$-th one) cannot be present in the linear
combination. So the $i$-th row of $B$ must be a
multiple of the $i$-th row of $C.$ But due to the
presence of the leading 1, the multiplying factor must
be $1.$ Hence the result.