# Why is row reduced echelon form of $A$ the same as that of $-A$?

I was reading a problem about linear transformations, which basically mapped a non-zero matrix to its row reduced echelon form and to $0$ otherwise. The question was whether the map was linear or not. I began by recalling that the row reduced echelon form for any given matrix was unique and so the linear combination of two matrices must be mapped to the linear combination of their echelon forms.

However, the author points out that the matrices $A$ and $-A$ have the same row reduced echelon form. To convince myself I programmed the Gaussian elimination algorithm and the result is indeed the same, except that the program outputs $-0$ in place of $0$. So I am guessing that this is indeed true. Could anyone, therefore, justify why the row reduced echelon form a matrix be the same as its additive inverse?

• Because you can row-reduce any row $k$ to row $-k$. – abiessu Oct 29 '17 at 21:49

Any two row-equivalent matrices will have the same reduced row-echelon form. The matrix $A$ is row-equivalent to $-A$, because multiplying a row by $-1$ is an elementary row operation. (Indeed, a slight generalization of this argument shows that $A$ and $kA$ will have the same RREF for any nonzero scalar $k$.)