Consider the set \begin{equation*} S=\set{\colvector{-7\\5\\1},\,\colvector{-6\\5\\0},\,\colvector{-12\\7\\4}} \end{equation*} from the vector space $\complex{3}$. Let $A$ be the matrix whose columns are the set $S$, and verify that $A$ is nonsingular. By Theorem NMLIC the elements of $S$ form a linearly independent set. Suppose that $\vect{b}\in\complex{3}$. Then $\linearsystem{A}{\vect{b}}$ has a (unique) solution (Theorem NMUS) and hence is consistent. By Theorem SLSLC, $\vect{b}\in\spn{S}$. Since $\vect{b}$ is arbitrary, this is enough to show that $\spn{S}=\complex{3}$, and therefore $S$ is a spanning set for $\complex{3}$ (Definition SSVS). (This set comes from the columns of the coefficient matrix of Archetype B.)
Now examine the situation for a particular choice of $\vect{b}$, say $\vect{b}=\colvector{-33\\24\\5}$. Because $S$ is a spanning set for $\complex{3}$, we know we can write $\vect{b}$ as a linear combination of the vectors in $S$, \begin{equation*} \colvector{-33\\24\\5}= (-3)\colvector{-7\\5\\1}+(5)\colvector{-6\\5\\0}+(2)\colvector{-12\\7\\4}. \end{equation*}
The nonsingularity of the matrix $A$ tells that the scalars in this linear combination are unique. More precisely, it is the linear independence of $S$ that provides the uniqueness. We will refer to the scalars $a_1=-3$, $a_2=5$, $a_3=2$ as a “representation of $\vect{b}$ relative to $S$.” In other words, once we settle on $S$ as a linearly independent set that spans $\complex{3}$, the vector $\vect{b}$ is recoverable just by knowing the scalars $a_1=-3$, $a_2=5$, $a_3=2$ (use these scalars in a linear combination of the vectors in $S$). This is all an illustration of the following important theorem, which we prove in the setting of a general vector space.