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?


Absolutely! A single counterexample is all you need. Nice work.
In general elements of $S$ will not have additive inverses in $S$. (Can you determine the exceptions?) Otherwise, the axioms are satisfied.
Geometrically speaking, I recommend that you focus on the lack of additive inverses. Note that if $A$ is a set of vectors such that every element of $A$ has an additive inverse in $A,$ then $A$ will be symmetric about the origin. That, in itself, won't be sufficient to make $A$ a vector subspace, but it will be necessary. Your set $S$ here is an octant of $3$-space. In general, an octant will not be a vector subspace, but a union of octants may be. (When?)
