Proving a Set is NOT a vector space Before I begin, I will emphasis I DO NOT want the full solution. I just want some hints.
Show that the set $S=\{\textbf{x}\in \mathbb{R}^3: x_{1} \leq 0$ and $x_{2}\geq 0 \}$ with the usual rules for addition and multiplication by a scalar in $\mathbb{R}^3$ is NOT  a vector space by showing that at least one of the vector space axioms is not satisfied. Give a geometric interpretation of the result.
My solution (so far): To show this, I will provide a counter example, I have selected axiom 6 (closure under multiplication of a scalar).
$\textbf{x} = \begin{pmatrix}x_{1}\\ x_{2}\\ x_{3}\end{pmatrix}$
Let $\lambda = -1, x_{1} = -2, x_{2} = 2, x_{3}=1$
$\lambda \textbf{x} = \lambda \begin{pmatrix}x_{1}\\ x_{2}\\ x_{3}\end{pmatrix}$ 
$= -1 \begin{pmatrix}-2\\ 2\\ 1\end{pmatrix}$
$= \begin{pmatrix}2\\ -2\\ -1\end{pmatrix}$
Clearly, as $\begin{pmatrix}2\\ -2\\ -1\end{pmatrix} \notin S$, as $x_{1} \nleqslant 0$ and $x_{2} \ngeqslant 0$ axiom (Multiplication by a scalar) does not hold. Hence $S$ is not a vector space.
My questions:


*

*Is my solution correct/reasoning? How can it be improved? (Please note I am new to Linear Algebra)

*Are there more axioms for which it doesn't hold besides the one I listed?

*It says to give a geometric interpretation of this result. I'm not sure how to go about doing this. Any hints?


The set of vectors violates the identity axiom.

The identity axiom: for each $\vec{v}\in{V}$, there is another element $\vec{w}\in{V}$ such that $\vec{v}+\vec{w}=\vec{0}$.

Any vector such that $v_1<0$ or $v_2>0$, its additive inverse would would have to have $w_1>0$, and $w_2<0$.
