[Math] How to all matrices have a _unique_ reduced row echelon form

linear algebramatrices

Wikipedia states (and I suspect the answer to my question could be that wikipedia should never have been my source) that

every matrix has a unique reduced row echelon form.

I am wondering how this can possibly be a unique matrix when any nonsingular matrix is row equivalent to the identity matrix, which is also their reduced row echelon form.

I am considering two possible interpretations that would result in Wikipedia's statement holding, which are (1) two row equivalent matrices are considered the same matrix$^1$ or (2) when Wikipedia says unique they mean only a single, rather than different from all others$^2$.

Because I am only a sophomore it seemed unwise for me to guess by myself which, if any, of these explanations would be the correct one, which is why I am once again asking for your help: what is meant when Wikipedia states "every matrix has a unique reduced row echelon form."


$^1$ if this were so than the conclusion that all matrices have a unique reduced row echelon form would be trivial since matrixes that have the same reduced row echelon form would just be defined as being the same matrix

$^2$ this would imply that they mean that for every matrix there exists one and only one reduced row echelon form, rather than two matrices have the same row echelon form if and only if they are the same matrix

Best Answer

I think that you have misinterpreted the statement. The statement "every matrix has a unique row-echelon form" can be restated as follows:

For every matrix $A$, there exists exactly one matrix $B$ such that $A$ is row-equivalent to $B$ and $B$ is in reduced row-echelon form (rref).

As an example, consider the matrices $$ A_1 = \pmatrix{1&2\\0&3}, \quad A_2 = \pmatrix{5&-1\\-1 & 7}, \quad I = \pmatrix{1&0\\0&1}. $$ Both $A_1$ and $A_2$ are invertible, so $I$ is the rref of both of these matrices. $I$ is the only matrix that is in rref and row-equivalent to $A_1$, so it is the rref of $A_1$. $I$ is also the only matrix that is in rref and row-equivalent to $A_2$, so it is the rref of $A_2$. The fact that $A_1$ and $A_2$ have the same rref does not contradict the fact that they have exactly one rref, i.e. "a unique" rref.