Answer
$S$ is not a subspace
Work Step by Step
$S$ is not a subspace because it is not closed under addition; indeed,
let $\left[\begin{array}{r}{-4}\\ {-2}\end{array}\right], \left[\begin{array}{r}{1}\\ {3}\end{array}\right]\in S$ and
$$\left[\begin{array}{r}{-4}\\ {-2}\end{array}\right]+\left[\begin{array}{r}{1}\\ {3}\end{array}\right]=\left[\begin{array}{r}{-3}\\ {1}\end{array}\right]\notin S.$$
