Closed Under Scalar Multiplication

Any scalar multiple of vector $\vec{x}\in S$ will also be in $S$ ($\vec{x}\in S,k\in R⟹k\vec{x}\in S$)

Example

Subset $S=\{(x,y,z)\in R^3;x\ge 0\}$ is not closed under scalar multiplication as: $(2,2,2)\in S$ but $-5(2,2,2)\notin S$

Proving Closed Under Scalar Muliplication