In Example RSC5 we used four vectors to create a span. With a relation of linear dependence in hand, we were able to “toss out” one of these four vectors and create the same span from a subset of just three vectors from the original set of four. We did have to take some care as to just which vector we tossed out. In the next example, we will be more methodical about just how we choose to eliminate vectors from a linearly dependent set while preserving a span.
Example COV deserves your careful attention, since this important example motivates the following very fundamental theorem.
Suppose that $S=\set{\vectorlist{v}{n}}$ is a set of column vectors. Define $W=\spn{S}$ and let $A$ be the matrix whose columns are the vectors from $S$. Let $B$ be the reduced row-echelon form of $A$, with $D=\set{\scalarlist{d}{r}}$ the set of indices for the pivot columns of $B$. Then
In Example COV, we tossed-out vectors one at a time. But in each instance, we rewrote the offending vector as a linear combination of those vectors with the column indices of the pivot columns of the reduced row-echelon form of the matrix of columns. In the proof of Theorem BS, we accomplish this reduction in one big step. In Example COV we arrived at a linearly independent set at exactly the same moment that we ran out of free variables to exploit. This was not a coincidence, it is the substance of our conclusion of linear independence in Theorem BS.
Here is a straightforward application of Theorem BS.