1. ## Vector spaces

I'm not too sure if this question belongs in this part of the forum.

Currently, I'm having a lot of trouble understanding this topic.

Show that the system $\displaystyle S$ with the usual rules for addition and multiplication by a scalar in $\displaystyle \mathbb{R}^3$, and where

$\displaystyle S=\{\mathbf{x}\in \mathbb{R}^3:2x_1+3x^3_2-4x^2_3=0\}$

is not a vector space by showing that at least one of the vector space axioms is not satisfied.

2. What ideas have you had so far?

3. Originally Posted by Ackbeet
What ideas have you had so far?
Well I know the zero vector is included. So it's non-empty. I'm a bit confused when showing closure under addition

So let $\displaystyle \mathbf{v,u}\in \mathbb{R}^3$

$\displaystyle 2v_1+3v_2^3-4v^2_3=0$ and $\displaystyle 2u_1+3u_2^3-4u^2_3=0$

$\displaystyle (2u_1+3u_2^3-4u^2_3)+(2v_1+3v_2^3-4v^2_3)=0+0$

What should I do next?

4. Well, addition is defined using the usual vector addition in $\displaystyle \mathbb{R}^{3}.$ That is, given two vectors

$\displaystyle \vec{r}_{1}=\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}$ and $\displaystyle \vec{r}_{2}=\begin{bmatrix}y_{1}\\ y_{2}\\ y_{3}\end{bmatrix},$ their sum is

$\displaystyle \vec{r}_{1}+\vec{r}_{2}=\begin{bmatrix}x_{1}+y_{1} \\ x_{2}+y_{2}\\ x_{3}+y_{3}\end{bmatrix}.$

I would try adding two vectors in S and seeing if the result looks like a vector in S. Where does that lead you?

5. Originally Posted by Ackbeet
Well, addition is defined using the usual vector addition in $\displaystyle \mathbb{R}^{3}.$ That is, given two vectors

$\displaystyle \vec{r}_{1}=\begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}$ and $\displaystyle \vec{r}_{2}=\begin{bmatrix}y_{1}\\ y_{2}\\ y_{3}\end{bmatrix},$ their sum is

$\displaystyle \vec{r}_{1}+\vec{r}_{2}=\begin{bmatrix}x_{1}+y_{1} \\ x_{2}+y_{2}\\ x_{3}+y_{3}\end{bmatrix}.$

I would try adding two vectors in S and seeing if the result looks like a vector in S. Where does that lead you?
What do you mean adding two vectors in s. Wouldn't you have to show that the sum of two vectors when substituted into the equation, it does not equal zero.

6. What do you mean adding two vectors in S?
Hmm. Yeah. I didn't explain that very clearly. Take two different vectors in S. Add them up using the vector addition I defined in Post # 4. See if the result is a vector in S. I would try this with a concrete example, which is all you need anyway if the result is not a vector in S. One counterexample is sufficient. Does this make sense? Can you move forward with this?

7. Sorry, its just that I'm having some trouble getting my head around this. Do you mean finding two vectors which satisfy the equation and actually adding them up?

For example, $\displaystyle (2,0,1)^T$ and adding it up with another vector and seeing whether it satisfies the equation? Is this a valid way of showing this?

8. Sure. $\displaystyle (2,0,1)^T$ is in the space. Can you come up with another vector in the space (I wouldn't recommend the zero vector, as it won't serve your purposes)?

9. Originally Posted by Ackbeet
Sure. $\displaystyle (2,0,1)^T$ is in the space. Can you come up with another vector in the space (I wouldn't recommend the zero vector, as it won't serve your purposes)?
Instead of vector addition, it wouldn't work for scalar multiplication, and therefore it is a not a vector space.

10. I agree, but you'd have to show the work. What do you get?

11. Anther vector would be $\displaystyle (-12,2,0)^T$ and when you add that with $\displaystyle (2,0,1)^T$ you get $\displaystyle (-10,2,1)^T$ it satisfies the set. So isn't it closed under addition.

For scalar multiplication couldn't we just let $\displaystyle \mathbf{x}=(2,0,1)^T$ and multiply that by a scalar 3. So the new vector would be $\displaystyle (6,0,3)^T$ which does not satisfy the equation. Therefore it is not closed under scalar multiplication

12. It satisfies the set. So isn't it closed under addition.
Actually, what that means is that you don't have a counterexample there, because that particular addition is closed!

Your scalar multiplication example works, and is sufficient to show that you don't have a vector space.

13. Originally Posted by Ackbeet
Actually, what that means is that you don't have a counterexample there, because that particular addition is closed!

Your scalar multiplication example works, and is sufficient to show that you don't have a vector space.
Thanks for the help!!

14. You're welcome. Have a good one!

15. ## non-planar space.

It is interesting to consider the question geometrically in a 3d coordinate system. Given x2 and x3, x1 is a point on a non-planar surface through the origin and x is a vector from the origin to a point on the surface. As such, ax and a(x+y) are not on the surface. Mathematically, if x and y satisfy the equation, then x + y, and ax and ay don't, ie, are not in the space as defined, as can be verified by substitution.

If the equation were linear, ie, the equation of a plane through the origin, then (x + y) and ax and ay would also be on the plane, ie, satisfy the linear equation.