# Thread: A couple of questions

1. ## A couple of questions

1. Let v1 = [-3; 3; -4], v2 = [3; -2; 8] and y = [-12; 9; h]

For what value of is in the plane spanned by and ?

2. Which of the following sets are subspaces of ?

A.
B. arbitrary number
C. arbitrary number
D. arbitrary number
E.
F.

Check all that apply.

Sort of desperate for help

2. Originally Posted by elven06

2. Which of the following sets are subspaces of ?

A.
B. arbitrary number
C. arbitrary number
D. arbitrary number
E.
F.

Check all that apply.

Sort of desperate for help

A. $\alpha=-1$
B. I think you can do this one.
C.
D.You try.
E. What about $\alpha=2$
F. $(1,0,0)+(0,1,0)$

3. Originally Posted by Drexel28
A. $\alpha=-1$
B. I think you can do this one.
C.
D.You try.
E. What about $\alpha=2$
F. $(1,0,0)+(0,1,0)$
I'm confused

4. Originally Posted by elven06
I'm confused
By what?

5. Originally Posted by Drexel28
By what?

6. Originally Posted by elven06
...which part?

7. Originally Posted by Drexel28
...which part?

I guess I still have no idea what to do in order to see which make it true... Your hints are tough to make use of given I have a weak grasp of the concept

8. Originally Posted by elven06
I guess I still have no idea what to do in order to see which make it true... Your hints are tough to make use of given I have a weak grasp of the concept
Ok. That's fine. So let's go through an example, ok?

A subspace must be closed under scalar multiplication and vector addition. So, for example F. is not a subspace since if we let $V=\left\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\right\}$ then surely $(1,0,0)\in V,(0,1,0)\in V$ but since $(1,0,0)+(0,1,0)=(1,1,0)$ has the quality that $1+1+0\ne 1$ we see that $(1,0,0)+(0,1,0)\notin V$. But, this HAS to happen in a subspace. So, $V$ is not a subspace.

9. Originally Posted by Drexel28
Ok. That's fine. So let's go through an example, ok?

A subspace must be closed under scalar multiplication and vector addition. So, for example F. is not a subspace since if we let $V=\left\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\right\}$ then surely $(1,0,0)\in V,(0,1,0)\in V$ but since $(1,0,0)+(0,1,0)=(1,1,0)$ has the quality that $1+1+0\ne 1$ we see that $(1,0,0)+(0,1,0)\notin V$. But, this HAS to happen in a subspace. So, $V$ is not a subspace.

Great, so I think I understand that B definitely is because you can multiply by any scalar to remain in R.

I'm thinking that C is also true because 0 is in R as well.

I'm not sure about others :/

10. Originally Posted by elven06
Great, so I think I understand that B definitely is because you can multiply by any scalar to remain in R.

I'm thinking that C is also true because 0 is in R as well.

I'm not sure about others :/
It has to satisfy ALL the properties of a subspace. Do you know them all?

11. Originally Posted by Drexel28
It has to satisfy ALL the properties of a subspace. Do you know them all?
I must not..