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 $C$.
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 $R$. 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) \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} \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} \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 $C$. 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 $C$, then compute the following linear combination using the definitions of the operations in $C$, \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) \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 $R$, so $C\subseteq\spn{R}$. 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 $a_2=-x+2y+1$.