1. Vector Spaces

The question:

Show that the set

$S = \{x \in \mathbb{R}^3 : x_1 \le 0, x_2 \ge 0 \}$

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 this result.

My attempt:

There's no solution to this answer in my text. My understanding is that this set is not closed under scalar multiplication, since multiplying by -1 will reverse the conditions on $x_1$ and $x_2$. Is this correct?

As for the second part of the question, it asks for a geometric interpretation, which I'm not sure about. It's a 3D object, but otherwise I don't know how else to interpret it. Any ideas?

Thanks.

2. Originally Posted by Glitch
The question:
My understanding is that this set is not closed under scalar multiplication, since multiplying by -1 will reverse the conditions on $x_1$ and $x_2$. Is this correct?
Yes, it is correct.

As for the second part of the question, it asks for a geometric interpretation, which I'm not sure about. It's a 3D object, but otherwise I don't know how else to interpret it. Any ideas?
Bear in mind that $S=A \times \mathbb{R}$ where $A$ is the second quadrant on the $x_1x_2$ plane.

Regards.

Fernando Revilla

3. Originally Posted by FernandoRevilla
Bear in mind that $S=A \times \mathbb{R}$ where $A$ is the second quadrant on the $x_1x_2$ plane.
I'm not sure what you mean by this. I'm unfamiliar with 'quadrants' in terms of planes in space (except for the standard Cartesian plane).

4. Yes, that's what he is talking about. The x and y axes divide the xy-plane into 4 quadrants. The usual convention is that the "first quadrant" is x> 0, y> 0, the "second quadrant" is x< 0, y> 0, the "third quadrant" is x< 0, y< 0, and the "fourth quadrant" is x> 0, y< 0. $\{(x, y)| x\le 0, y\ge 0\}$ is the second quadrant, together with the axes bounding it.

In fact, that can be extended to three dimensions: the x, y, and z coordinate planes divide the space into 8 "octants". Just take the four quadrants above with z> 0 to get four of them and take the four quadrants above with z< 0 to get the other four.

$S= \{(x_1, x_2, x_3)|x_1\le 0, x_2\ge 0\}$ is the second and sixth octants, together with the planes bounding them.

5. Ahh that makes more sense. Thank you!