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# 18c More commentary on a spanning set -- On ordering.
By now you learned how to remove redundant column vectors in a spanning set using our algorithm procedure. But notice the resulting spanning set that we get largely **depends on the order to which we list those column vectors!** Despite this, **the number of pivoted vectors left to keep will always be the same**, which is an amazing fact.
Let us illustrate this. Consider the set $$
S = \operatorname{span} (\begin{bmatrix}1\\2\\3\end{bmatrix}
, \begin{bmatrix}4\\5\\6\end{bmatrix}
, \begin{bmatrix}7\\8\\9\end{bmatrix}
),
$$
if we use our algorithm to produce a more economical spanning set, we may get something like this:$$
\begin{bmatrix}1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9\end{bmatrix} \stackrel{\text{row}}\sim \begin{bmatrix}1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 0\end{bmatrix}.
$$So we conclude $$
S = \operatorname{span} (\begin{bmatrix}1\\2\\3\end{bmatrix}
, \begin{bmatrix}4\\5\\6\end{bmatrix}
, \cancel{\begin{bmatrix}7\\8\\9\end{bmatrix}}
)= \operatorname{span} (\begin{bmatrix}1\\2\\3\end{bmatrix}
, \begin{bmatrix}4\\5\\6\end{bmatrix}
).
$$
But wait, you shout: The same set $S$ can be written as $$
S = \operatorname{span} (\begin{bmatrix}1\\2\\3\end{bmatrix}
, \begin{bmatrix}4\\5\\6\end{bmatrix}
, \begin{bmatrix}7\\8\\9\end{bmatrix}
)=\operatorname{span} (
\begin{bmatrix}4\\5\\6\end{bmatrix}
, \begin{bmatrix}7\\8\\9\end{bmatrix}
,\begin{bmatrix}1\\2\\3\end{bmatrix}
)
$$as we merely shuffle around **the ordering** of the spanning set. The resulting span is still the same! They produce the same collection of linear combinations. But if we use this new ordering, we see that $$
\begin{bmatrix}4 & 7 & 1 \\
5 & 8 & 2 \\
6 & 9 & 3\end{bmatrix} \stackrel{\text{row}}\sim \begin{bmatrix}1 & 0 & 2 \\
0 & 1 & -1 \\
0 & 0 & 0\end{bmatrix},
$$so we conclude $$
S = \operatorname{span} (\begin{bmatrix}4\\5\\6\end{bmatrix}
,\begin{bmatrix}7\\8\\9\end{bmatrix}
,\cancel{\begin{bmatrix}1\\2\\3\end{bmatrix}})
= \operatorname{span} (\begin{bmatrix}4\\5\\6\end{bmatrix}
,\begin{bmatrix}7\\8\\9\end{bmatrix}).
$$In both cases, we reduced the spanning set to just $2$ vectors for the set $S$, although the one we got rid of is different. Again, spanning sets are not unique, and using our algorithm we can preferentially keep what vectors we like to be in the final spanning set by putting it in the front of the list.
Later, ordering of things will be of importance (and then much later, not so important again).