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I know $Ax=b$ has a solution iff $b$ is in the column space of $A$. In this case, if $A$ columns are $a_1,...,a_n$ and $x's$ components are $x_1,...,x_n$ then these components satisfy the equation $a_1x_1+...+a_nx_n = b$ (i.e. $x$'s components are the necessary weights, or coordinates of $b$ in the basis formed from $A$ columns). But given that elemntary row operations do chagne the column space, how come they don't change the solution set (i.e., if rref($A$) columns are $a_1',...,a_n'$ and the elemntary row operations change b to $b'$, then the same vector $\vec{x}$ satisfies ($1$) $b=a_1x_1+...+a_nx_n$, (2) $b'=a_1'x_1+...+a_n'x_n$?

I understand the algebric proof for the statement that elemntary row operations don't change the solution set, but I would appreciate a geometric intuition. I guess it's somewhat related to the fact the elementary row operation $Ri$-->$Ri+cRj$ doesn't change the volume of the parallelepiped formed by $A's$ columns, but I fail to see exactly how.

Thanks.

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The thing to realize is that if you have any two vectors $b_1, b_2$ then the following things hold for spans: $\langle b_1, b_2\rangle=\langle b_2, b_1\rangle$, and $\langle b_1\rangle=\langle \lambda b_1\rangle$ for any nonzero scalar $\lambda$, and $\langle b_1, b_2\rangle=\langle b_1+b_2, b_2\rangle$. These correspond to elementary row operations.

Thus, doing any of these operations inside a larger set of vectors $X$ does not change the span.

If you don't change the span, then you certainly also don't change the orthogonal space (= the solution set), since $\{y\mid xy^\top = 0 \text{ for all } x\in span(X)\}$ only depends on the span of $X$, and not the particular things used to generate it.

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    $\begingroup$ Thanks for your explanation! It certainly made things clearer. But I wonder if there is also a more concrete geometric way to look at the process. Is there a way to connect this to the usual geometric interpretation of row reduction of an invertible matrix - mapping the parallelepiped made by $A$'s columns back to the unit square? what happens to $b$ in this process? $\endgroup$ Commented Sep 22, 2016 at 15:09

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