## Elementary Linear Algebra 7th Edition

$S$ does not span $R^2$.
The set $S=\{(-1,2)\}$ is not a basis for $R^2$ because $S$ does not span $R^2$. For example, suppose the vector $(-1,0)\in R^2$ and writing it as a linear combination of the elements o $S$, we get $$(-1,0)=c(-1,2), \quad c\in R.$$ Which yields the following system of equations $$-c=-1, \quad 2c=0.$$ It is easy to note that the above system is in-consistent, since $c=1$ the solution of the first equation does not satisfy the second equation.