Uniqueness of reduced row echelon form

Here is a short proof that uses only the row space. For a vector subspace $W\subseteq {\mathbb{F}}^n$, let $d_j$ be the dimension of its projection map onto the first $j$ coordinates, with $d_0=0$. In any reduced row echelon form with row space $W$, $d_j$ counts the pivot columns among those first $j$ columns: later-pivot rows project to zero, while earlier-pivot rows remain linearly independent. Thus the positions where $d_j-d_{j-1}=1$ determine all pivot positions from $W$ alone.

If those positions are $p_1,\dots ,p_r$, the map sending $w\in W$ to $(w_{p_1},\dots ,w_{p_r})$ is an isomorphism $W\to {\mathbb{F}}^r$: the reduced nonzero rows map to the standard basis. Its $i$th row is therefore the unique vector of $W$ with pivot-coordinate values equal to the $i$th standard basis vector. Any other reduced row echelon form with that row space has identical nonzero rows in identical order. At a fixed matrix size, the remaining rows must be zero. Together with existence by Gaussian elimination, this proves uniqueness.