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?

 A: *

*Yes, your reasoning is correct. Before I read your solution, this would be how I would have done it too. If you want to write down the solutoin I would probably write it like this:

Note that $v = (1,1,1) \in S$. If $S$ is a vector space then $-1v$ would be in $S$. But $-1(1,1,1) = (-1,-1,-1)$ is not in $S$ because the first coordinate is not non-negative.



*I don't see any other axioms that $S$ doesn't satisfy. All other ways of saying that $S$ is not a vector space seems to me to come down to what you have.


*Now what you have proves is that the set is not closed under scalar multiplication. This means that the set $S$ doesn't contain all lines. Try to think about how $S$ looks like. You have all points $(x,y,z)$ in $\mathbb{R}^3$ with $x$ and $y$ non negative. Now try to draw lines through the origin.
A: Hint
To see that $S$ isn't a vector space by an other method select two vectors $x,y\in S$ such that $x-y\not\in S$. How we can choose the components of $x$ and $y$ to find the desired result.
A: Your answer is correct.
All Vector-space axioms are:
$$(V1) 0\in S \qquad (\checkmark)$$
$$(V2) a+b \in S \qquad \forall a,b\in S \quad (\checkmark)$$
$$(V3) \lambda a \in S \qquad \forall a\in S, \lambda \in K \quad (\text{f})$$
Also $+$ must be associative and commutative and $\lambda (a+b) = \lambda a + \lambda b$ (distributive)
A: *

*It's correct. Personally, I'd use a simpler example, i.e., $e_1 = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}^T$.


*What would be the additive neutral element? And additive inverse of any vector?


*A subspace is a plane (not necessarily 2D) through $0$. Since $0 \in S$, this is obviously not a plane, but its part. Look more closely: $S$ is a...

 ...quadrant.

A: 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?)
A: 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 $w_2<0$.
