Proof of uniqueness of reduced row echelon form I've found a proof of uniqueness of reduced row echelon form. I have certian doubts with regard to this sentence: "It follows that R' and S' are (row) equivalent since deletion of columns does not affect row equivalence, and that they are reduced but not equal."
But wait a second, we have chosen R' and S' matrices such that they are identical in all but the last column. So the matrices are NOT row equivalent. Could anyone explain what I'm missing here?
 A: The way it's written is kind of confusing and I think incorrect, but I also think the main idea works. First of all, your question:
$R$ and $S$ are row-equivalent because they both come from row-reducing the same matrix (call this original matrix $A$). Thus, $R'$ and $S'$ are row-equivalent because they can be seen as coming from a matrix $A'$ which is defined by deleting exactly the same columns that you deleted from $R'$ (or $S'$) via the same sets of row operations. 
The confusing part comes in at the end. The possibilities are: 
(i) the augmented systems are both uniquely solvable, which implies that $r' = s'$ (implying that $R'$ = $S'$, which is a contradiction);
(ii) $R'$ is consistent but $S'$ is not (or vice versa), which happens when $R$ and $S$ first disagree at a place where the column of one matrix is a pivot but the corresponding column of the other matrix isn't. But this is also a contradiction since $R'$ and $S'$ are row-equivalent (and thus share the same solution set).
The pdf refers to a possibility that "both $R'$ and $S'$ are inconsistent" but I don't think that makes sense, because of the way $R'$ and $S'$ are defined-- you wouldn't have chosen those columns to define $R'$ and $S'$ in the first place since they are obviously equal. The only 'interesting' choices are the two above.
