Is the converse of this subspace statement true?

Question

If two vectors are in a subspace, then so is their vector sum. Is the converse of this statement true?

Answer 1

No. Take $\mathbb{R}^2$ and consider the subpace $S:= \{(x,x) : x \in \mathbb{R} \}$. Then $(1,0) + (0,1) \in S$ but neither summand lies in $S$.

Answer 2

No. $$(1,0)+(0,1)=(1,1)\in \text{span}\big((1,1)\big)$$ but neither are in the subspace themselves. $$(1,0),(0,1)\notin \text{span}\big((1,1)\big)$$

Answer 3

Given an arbitrary vector space $V\neq\{0\}$, take the subspace $\{0\}$, then for any nonzero $v \in V$, we have $v, -v \not\in \{0\}$ but $v+ -v \in \{0\}$.