### SectionSSSpanning Sets

In this section we will provide an extremely compact way to describe an infinite set of vectors, making use of linear combinations. This will give us a convenient way to describe the solution set of a linear system, the null space of a matrix, and many other sets of vectors.

#### SubsectionSSVSpan of a Set of Vectors

In Example VFSAL we saw the solution set of a homogeneous system described as all possible linear combinations of two particular vectors. This is a useful way to construct or describe infinite sets of vectors, so we encapsulate the idea in a definition.

##### Definition SSCV Span of a Set of Column Vectors

Given a set of vectors $S=\{\vectorlist{u}{p}\}$, their span, $\spn{S}$, is the set of all possible linear combinations of $\vectorlist{u}{p}$. Symbolically, \begin{align*} \spn{S}&=\setparts{\lincombo{\alpha}{u}{p}}{\alpha_i\in\complex{\null},\,1\leq i\leq p}\\ &=\setparts{\sum_{i=1}^{p}\alpha_i\vect{u}_i}{\alpha_i\in\complex{\null},\,1\leq i\leq p} \end{align*}

The span is just a set of vectors, though in all but one situation it is an infinite set. (Just when is it not infinite?) So we start with a finite collection of vectors $S$ ($p$ of them to be precise), and use this finite set to describe an infinite set of vectors, $\spn{S}$. Confusing the finite set $S$ with the infinite set $\spn{S}$ is one of the most persistent problems in understanding introductory linear algebra. We will see this construction repeatedly, so let us work through some examples to get comfortable with it. The most obvious question about a set is if a particular item of the correct type is in the set, or not in the set.

##### Example SCAASpan of the columns of Archetype A

Having analyzed Archetype A in Example SCAA, we will of course subject Archetype B to a similar investigation.

#### SubsectionSSNSSpanning Sets of Null Spaces

We saw in Example VFSAL that when a system of equations is homogeneous the solution set can be expressed in the form described by Theorem VFSLS where the vector $\vect{c}$ is the zero vector. We can essentially ignore this vector, so that the remainder of the typical expression for a solution looks like an arbitrary linear combination, where the scalars are the free variables and the vectors are $\vectorlist{u}{n-r}$. Which sounds a lot like a span. This is the substance of the next theorem.

##### Theorem SSNS Spanning Sets for Null Spaces

Suppose that $A$ is an $m\times n$ matrix, and $B$ is a row-equivalent matrix in reduced row-echelon form. Suppose that $B$ has $r$ pivot columns, with indices given by $D=\set{d_1,\,d_2,\,d_3,\,\ldots,\,d_r}$, while the $n-r$ non-pivot columns have indices $F=\set{f_1,\,f_2,\,f_3,\,\ldots,\,f_{n-r},\,n+1}$. Construct the $n-r$ vectors $\vect{z}_j$, $1\leq j\leq n-r$ of size $n$, \begin{equation*} \vectorentry{\vect{z}_j}{i}= \begin{cases} 1&\text{if $i\in F$, $i=f_j$}\\ 0&\text{if $i\in F$, $i\neq f_j$}\\ -\matrixentry{B}{k,f_j}&\text{if $i\in D$, $i=d_k$} \end{cases} \end{equation*}

Then the null space of $A$ is given by \begin{equation*} \nsp{A}=\spn{\left\{\vectorlist{z}{n-r}\right\}} \end{equation*}

Notice that the hypotheses of Theorem VFSLS and Theorem SSNS are slightly different. In the former, $B$ is the row-reduced version of an augmented matrix of a linear system, while in the latter, $B$ is the row-reduced version of an arbitrary matrix. Understanding this subtlety now will avoid confusion later.

##### Sage SSNSSpanning Sets for Null Spaces

Here is an example that will simultaneously exercise the span construction and Theorem SSNS, while also pointing the way to the next section.