[Math] Proving a Set is NOT a vector space

linear algebravector-spaces

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?

Best Answer

  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.

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