Begin with the finite set of three vectors of size \(3\)
\begin{equation*}
S=\{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3\}
=\left\{
\colvector{1\\2\\1},\,\colvector{-1\\1\\1},\,\colvector{2\\1\\0}
\right\}
\end{equation*}
and consider the infinite set \(\spn{S}\text{.}\) The vectors of \(S\) could have been chosen to be anything, but for reasons that will become clear later, we have chosen the three columns of the coefficient matrix in Archetype A.

First, as an example, note that
\begin{equation*}
\vect{v}=(5)\colvector{1\\2\\1}+(-3)\colvector{-1\\1\\1}+(7)\colvector{2\\1\\0}=\colvector{22\\14\\2}
\end{equation*}
is in \(\spn{S}\text{,}\) since it is a linear combination of \(\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3\text{.}\) We write this succinctly as \(\vect{v}\in\spn{S}\text{.}\) There is nothing magical about the scalars \(\alpha_1=5,\,\alpha_2=-3,\,\alpha_3=7\text{,}\) they could have been chosen to be anything. So repeat this part of the example yourself, using different values of \(\alpha_1,\,\alpha_2,\,\alpha_3\text{.}\) What happens if you choose all three scalars to be zero?

So we know how to quickly construct sample elements of the set \(\spn{S}\text{.}\) A slightly different question arises when you are handed a vector of the correct size and asked if it is an element of \(\spn{S}\text{.}\) For example, is \(\vect{w}=\colvector{1\\8\\5}\) in \(\spn{S}\text{?}\) More succinctly, \(\vect{w}\in\spn{S}\text{?}\)

To answer this question, we will look for scalars \(\alpha_1,\,\alpha_2,\,\alpha_3\) so that
\begin{align*}
\alpha_1\vect{u}_1+\alpha_2\vect{u}_2+\alpha_3\vect{u}_3&=\vect{w}\\
\end{align*}
By Theorem SLSLC solutions to this vector equation are solutions to the system of equations
\begin{align*}
\alpha_1-\alpha_2+2\alpha_3&=1\\
2\alpha_1+\alpha_2+\alpha_3&=8\\
\alpha_1+\alpha_2&=5\\
\end{align*}
Building the augmented matrix for this linear system, and row-reducing, gives
\begin{align*}
\begin{bmatrix}
\leading{1} & 0 & 1 & 3\\
0 & \leading{1} & -1 & 2\\
0 & 0 & 0 & 0
\end{bmatrix}\text{.}
\end{align*}

This system has infinitely many solutions (there is a free variable in \(x_3\)), but all we need is one solution vector. The solution \(\alpha_1 = 2\text{,}\) \(\alpha_2 = 3\text{,}\) \(\alpha_3 = 1\text{,}\) tells us that
\begin{equation*}
(2)\vect{u}_1+(3)\vect{u}_2+(1)\vect{u}_3=\vect{w}
\end{equation*}
so we are convinced that \(\vect{w}\) really is in \(\spn{S}\text{.}\) Notice that there are an infinite number of ways to answer this question affirmatively.

We could choose a different solution, this time choosing the free variable to be zero. Then \(\alpha_1 = 3\text{,}\) \(\alpha_2 = 2\text{,}\) \(\alpha_3 = 0\text{,}\) shows us that
\begin{equation*}
(3)\vect{u}_1+(2)\vect{u}_2+(0)\vect{u}_3=\vect{w}\text{.}
\end{equation*}

Verifying the arithmetic in this second solution will make it obvious that \(\vect{w}\) is in this span. And of course, we now realize that there are an infinite number of ways to realize \(\vect{w}\) as element of \(\spn{S}\text{.}\)

Let us ask the same type of question again, but this time with \(\vect{y}=\colvector{2\\4\\3}\text{,}\) i.e. is \(\vect{y}\in\spn{S}\text{?}\)

So we will look for scalars \(\alpha_1,\,\alpha_2,\,\alpha_3\) so that
\begin{align*}
\alpha_1\vect{u}_1+\alpha_2\vect{u}_2+\alpha_3\vect{u}_3&=\vect{y}\\
\end{align*}
By Theorem SLSLC solutions to this vector equation are the solutions to the system of equations
\begin{align*}
\alpha_1-\alpha_2+2\alpha_3&=2\\
2\alpha_1+\alpha_2+\alpha_3&=4\\
\alpha_1+\alpha_2&=3\\
\end{align*}
Building the augmented matrix for this linear system, and row-reducing, gives
\begin{align*}
\begin{bmatrix}
\leading{1} & 0 & 1 & 0\\
0 & \leading{1} & -1 & 0\\
0 & 0 & 0 & \leading{1}
\end{bmatrix}\text{.}
\end{align*}

This system is inconsistent (there is a pivot column in the last column, Theorem RCLS), so there are no scalars \(\alpha_1,\,\alpha_2,\,\alpha_3\) that will create a linear combination of \(\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3\) that equals \(\vect{y}\text{.}\) More precisely, \(\vect{y}\not\in\spn{S}\text{.}\)

There are three things to observe in this example. (1) It is easy to construct vectors in \(\spn{S}\text{.}\) (2) It is possible that some vectors are in \(\spn{S}\) (e.g. \(\vect{w}\)), while others are not (e.g. \(\vect{y}\)). (3) Deciding if a given vector is in \(\spn{S}\) leads to solving a linear system of equations and asking if the system is consistent.

With a computer program in hand to solve systems of linear equations, could you create a program to decide if a vector was, or was not, in the span of a given set of vectors? Is this art or science?

This example was built on vectors from the columns of the coefficient matrix of Archetype A. Study the determination that \(\vect{v}\in\spn{S}\) and see if you can connect it with some of the other properties of Archetype A.