In Example LIC we determined that the set \(R=\set{(1,\,0),\,(6,\,3)}\) is linearly independent in the crazy vector space \(C\) (Example CVS). We now show that \(R\) is a spanning set for \(C\text{.}\)

Given an arbitrary vector \((x,\,y)\in C\) we desire to show that it can be written as a linear combination of the elements of \(R\text{.}\) In other words, are there scalars \(a_1\) and \(a_2\) so that
\begin{equation*}
(x,\,y)=a_1(1,\,0) + a_2(6,\,3)\text{.}
\end{equation*}

We will act as if this equation is true and try to determine just what \(a_1\) and \(a_2\) would be (as functions of \(x\) and \(y\)). Recall that our vector space operations are unconventional and are defined in Example CVS.
\begin{align*}
(x,\,y)&=a_1(1,\,0) + a_2(6,\,3)\\
&= (1a_1+a_1-1,\,0a_1+a_1-1) + (6a_2+a_2-1,\,3a_2+a_2-1)\\
&= (2a_1-1,\,a_1-1) + (7a_2-1,\,4a_2-1)\\
&= (2a_1-1+7a_2-1+1,\,a_1-1+4a_2-1+1)\\
&= (2a_1+7a_2-1,\,a_1+4a_2-1)
\end{align*}

Equality in \(C\) then yields the two equations,
\begin{align*}
2a_1+7a_2-1&=x\\
a_1+4a_2-1&=y
\end{align*}
which becomes the linear system with a matrix representation
\begin{equation*}
\begin{bmatrix}
2 & 7 \\ 1 & 4
\end{bmatrix}
\colvector{a_1\\a_2}
=
\colvector{x+1\\y+1}\text{.}
\end{equation*}

The coefficient matrix of this system is nonsingular, hence invertible (Theorem NI), and we can employ its inverse to find a solution (Theorem TTMI, Theorem SNCM),
\begin{equation*}
\colvector{a_1\\a_2}=
\inverse{\begin{bmatrix} 2 & 7 \\ 1 & 4 \end{bmatrix}}\colvector{x+1\\y+1}=
\begin{bmatrix} 4 & -7 \\ -1 & 2 \end{bmatrix}\colvector{x+1\\y+1}=
\colvector{4x-7y-3\\-x+2y+1}\text{.}
\end{equation*}

We could chase through the above implications backwards and take the existence of these solutions as sufficient evidence for \(R\) being a spanning set for \(C\text{.}\) Instead, let us view the above as simply scratchwork and now get serious with a simple direct proof that \(R\) is a spanning set. Ready? Suppose \((x,\,y)\) is any vector from \(C\text{,}\) then compute the following linear combination using the definitions of the operations in \(C\text{,}\)
\begin{align*}
(4x&-7y-3)(1,\,0)+(-x+2y+1)(6,\,3)\\
&=\left(1(4x-7y-3)+(4x-7y-3)-1,\,0(4x-7y-3)+(4x-7y-3)-1\right)+\\
&\quad\left(6(-x+2y+1)+(-x+2y+1)-1,\,3(-x+2y+1)+(-x+2y+1)-1\right)\\
&=(8x-14y-7,\,4x-7y-4)+(-7x+14y+6,\,-4x+8y+3)\\
&=((8x-14y-7)+(-7x+14y+6)+1,\,(4x-7y-4)+(-4x+8y+3)+1)\\
&=(x,\,y)\text{.}
\end{align*}

This final sequence of computations in \(C\) is sufficient to demonstrate that any element of \(C\) *can* be written (or expressed) as a linear combination of the two vectors in \(R\text{,}\) so \(C\subseteq\spn{R}\text{.}\) Since the reverse inclusion \(\spn{R}\subseteq C\) is trivially true, \(C=\spn{R}\) and we say \(R\) spans \(C\) (Definition SSVS). Notice that this demonstration is no more or less valid if we hide from the reader our scratchwork that suggested \(a_1=4x-7y-3\) and \(a_2=-x+2y+1\text{.}\)