### Section LDS Linear Dependence and Spans

In any linearly dependent set there is always one vector that can be written as a linear combination of the others. This is the substance of the upcoming Theorem DLDS. Perhaps this will explain the use of the word “dependent.” In a linearly dependent set, at least one vector “depends” on the others (via a linear combination).

Indeed, because Theorem DLDS is an equivalence (Proof Technique E) some authors use this condition as a definition (Proof Technique D) of linear dependence. Then linear independence is defined as the logical opposite of linear dependence. Of course, we have *chosen* to take Definition LICV as our definition, and then follow with Theorem DLDS as a theorem.

#### Subsection LDSS Linearly Dependent Sets and Spans

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.

##### Theorem DLDS Dependency in Linearly Dependent Sets

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.

##### Example RSC5 Reducing a span in $\complex{5}$

##### Sage RLD Relations of Linear Dependence

#### Subsection COV Casting Out Vectors

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 Casting out vectors

##### Sage COV Casting Out Vectors

Example COV deserves your careful attention, since this important example motivates the following very fundamental theorem.

##### Theorem BS Basis of a Span

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

- $T=\set{\vect{v}_{d_1},\,\vect{v}_{d_2},\,\vect{v}_{d_3},\,\ldots\,\vect{v}_{d_r}}$ is a linearly independent set.
- $W=\spn{T}$.

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.