Let $L$ denote a linear space. Let $x_1,x_2,\cdots,x_n \in L$. Observe that
$span\{x_1,x_2,\cdots,x_n\} = span\{x_2,x_1,\cdots,x_n\}$
This motivates the following question: under what operations with $x_1,x_2,\cdots,x_n$ the $span\{x_1,x_2,\cdots,x_n\}$ remains unchanged, i.e. is invariant?
Observe that $span\{x_1,x_2,\cdots,x_n\}=span\{x_1 + x_2,x_2,\cdots,x_n\}$
Definition. We say that $\{y_1,y_2,\cdots,y_n\}$ is obtained from $\{x_1,x_2,\cdots,x_n\}$ using elementary operations and write $\{x_1,x_2,\cdots,x_n\}∼\{y_1,y_2,\cdots,y_n\}$ if one of the following holds:
(summation) $y_i=x_i+x_j$ for some $i$ and $j$, and $y_k=x_k$ for $k\neq i$, or
(multiplication with a nonzero number) $y_i=\alpha x_i$ for some $i$ and some number $\alpha \neq 0$, and $y_k=x_k$ for $k\neq i$, or
(interchanging two vectors) $y_i = x_j$, $y_j=x_i$ for some $i$ and $j$, and $y_k=x_k$ for $k\neq i$, or
$\{y_1,y_2,\cdots,y_n\}$ is obtained from $\{x_1,x_2,\cdots,x_n\}$ using finitely many steps 1, 2 or 3.
So my question is: what is this condition $y_k=x_k$ for $k\neq i$ and why is it needed?