# row space of A is equal to the row space of rref?

This is a proof from a textbook

What I don't undersdand is, clearly the cofficients for r_i is not equal, unless a_j is 0 (k has to be non-zero), but we want a_j to be any number, don't we? so a_i is not equal to a_i + k*a_j, which means vector whose a_j is non-zero in the original spanning sets cannot be a linear combination of the new spanning sets, then how can we say row operations do not change the row space of a matrix?

• Given any linear combination of $r_1,\dots,(r_j+kr_i),\dots,r_n$, this can be written as a linear combination of $r_1,\dots,r_n$ by moving the term $a_j k r_i$ and grouping it with the other $r_i$ term. That's all that matters, the fact that $a_i+ka_j$ isn't the same as $a_i$ doesn't matter. – Ian Apr 24 '16 at 2:26