If we use a linearly dependent set to construct a span, then we can always create the same infinite set with a starting set that is one vector smaller in size. We will illustrate this behavior in Example RSC5. However, this will not be possible if we build a span from a linearly independent set. So in a certain sense, using a linearly independent set to formulate a span is the best possible way — there are not any extra vectors being used to build up all the necessary linear combinations. OK, here is the theorem, and then the example.
Suppose that $S=\set{\vectorlist{u}{n}}$ is a set of vectors. Then $S$ is a linearly dependent set if and only if there is an index $t$, $1\leq t\leq n$ such that $\vect{u_t}$ is a linear combination of the vectors $\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3,\,\ldots,\,\vect{u}_{t-1},\,\vect{u}_{t+1},\,\ldots,\,\vect{u}_n$.
This theorem can be used, sometimes repeatedly, to whittle down the size of a set of vectors used in a span construction. We have seen some of this already in Example SCAD, but in the next example we will detail some of the subtleties.