# 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:

1. Is my solution correct/reasoning? How can it be improved? (Please note I am new to Linear Algebra)
2. Are there more axioms for which it doesn't hold besides the one I listed?
3. It says to give a geometric interpretation of this result. I'm not sure how to go about doing this. Any hints?
• You are correct. – Emily Aug 6 '13 at 14:24
• 3. If I give you a point in $\mathbb{R}^3$ and tell you that it belongs to some subspace, what geometric object must that subspace contain? Does $S$ contain this geometric object? – RghtHndSd Aug 6 '13 at 14:25
• @rghthndsd $S$ contains $\langle e_3 \rangle$ ;-) – AlexR Aug 6 '13 at 14:31
• @Alex: You seem to be suggesting my comment is wrong. While it is true that $S$ contains $\langle e_3 \rangle$ , I think you're confusing "there exists" with "for all". – RghtHndSd Aug 6 '13 at 16:05
• @rghthndsd You didn't clarify "for all". You just said "If I gave you one point". If that is valid for all, it still needn't bee a subspace; consider $\langle e_1 \rangle \cup \langle e_2 \rangle$, which contains any linear hull of one element. Of course it is necessary for a subspace to contain the linear hull of any of its elements, but this is just what $(V3)$ states (refering to my answer), and what the OP did prove goes wrong for a special vector. – AlexR Aug 6 '13 at 16:10

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.

1. 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.

2. 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.

3. 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.

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

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

3. 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...

• The additive neutral element would be the usual one. – Cameron Buie Aug 6 '13 at 14:28
• Exactly, and that one is in $S$. That's why I also asked about the additive inverse. – Vedran Šego Aug 6 '13 at 14:29
• Ah. You were just mentioning something to check in general, not suggesting an axiom which failed in this case. Gotcha. – Cameron Buie Aug 6 '13 at 14:30
• Yes, it's a way to look at these things. "Neutral... check. Inverse... whoops!" I'm sorry I wasn't clear enough. – Vedran Šego Aug 6 '13 at 14:31

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)

• Accidentally sent before complete... sorry for that. – AlexR Aug 6 '13 at 14:26

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?)

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 $w_2<0$.