Is the set $R=\set{(1,\,0),\,(6,\,3)}$ linearly independent in the crazy vector space $C$ (Example CVS)?
We begin with an arbitrary relation of linear dependence on $R$ \begin{align*} \zerovector &= a_1(1,\,0) + a_2(6,\,3)&&\knowl{./knowls/definition.RLD.knowl}{\text{Definition RLD}}\\ \end{align*} and then massage it to a point where we can apply the definition of equality in $C$. Recall the definitions of vector addition and scalar multiplication in $C$ are not what you would expect. \begin{align*} (&-1,\,-1)\\ &=\zerovector &&\knowl{./knowls/example.CVS.knowl}{\text{Example CVS}}\\ &=a_1(1,\,0) + a_2(6,\,3) &&\knowl{./knowls/definition.RLD.knowl}{\text{Definition RLD}}\\ &=(1a_1+a_1-1,\,0a_1+a_1-1) + (6a_2+a_2-1,\,3a_2+a_2-1) &&\knowl{./knowls/example.CVS.knowl}{\text{Example CVS}}\\ &=(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) &&\knowl{./knowls/example.CVS.knowl}{\text{Example CVS}}\\ &=(2a_1+7a_2-1,\,a_1+4a_2-1) \end{align*}
Equality in $C$ (Example CVS) then yields the two equations, \begin{align*} 2a_1+7a_2-1&=-1\\ a_1+4a_2-1&=-1 \end{align*} which becomes the homogeneous system \begin{align*} 2a_1+7a_2&=0\\ a_1+4a_2&=0 \end{align*}
Since the coefficient matrix of this system is nonsingular (check this!) the system has only the trivial solution $a_1=a_2=0$. By Definition LI the set $R$ is linearly independent. Notice that even though the zero vector of $C$ is not what we might have first suspected, a question about linear independence still concludes with a question about a homogeneous system of equations. Hmmm.