Chapter 3 - Matrices - 3.5 Subspaces, Basis, Dimension, and Rank - Exercises 3.5 - Page 209: 8

$S$ is not a subspace

$S$ is not a subspace because it is not closed under addition. Let $u=\left[\begin{array}{r}{2} \\ {0}\\{2}\end{array}\right], v=\left[\begin{array}{r}{2} \\ {0}\\{-2}\end{array}\right]\in S$ but $$u+v=\left[\begin{array}{r}{2} \\ {0}\\{2}\end{array}\right]+\left[\begin{array}{r}{2} \\ {0}\\{-2}\end{array}\right]=\left[\begin{array}{r}{2} \\ {0}\\{0}\end{array}\right]\notin S.$$