# Homework Help: Linear Subspace in R^5 and spanning vectors

1. Nov 9, 2009

### KaiserBrandon

1. The problem statement, all variables and given/known data
Denote by W the set of all vectors that are of the form x = (a, b 2a, 3b,-a), in which a
and b are arbitrary real numbers. Show that W is a linear subspace of R^5. Also find a
set of spanning vectors for W. What kind of geometric object is W?

2. Relevant equations

none

3. The attempt at a solution

So I found that W is a linear subspace of R^5 since it meets the three criteria. For the set of spanning vectors, I know that by definition that these are the set of all linear combination of these vectors which satisfy the form x. However, I don't really know how to denote these vectors.

2. Nov 9, 2009

### Staff: Mentor

x = (a, b, 2a, 3b,-a), so
x1=.1a
x2=.........1b
x3=.2a
x4=.........3b
x5=-1a

This is kind of a hint.

3. Nov 9, 2009

### lanedance

so how many dimensions is your subsapce? you will need that amount of spanning vectors... consider how many parameters are variable, and and easy way to make linearly independent vectors from your parmeters... (choose a & b values)

Last edited: Nov 9, 2009
4. Nov 9, 2009

### KaiserBrandon

alright, I'm just gonna take a stab at this. would the spanning vectors be x=2a+4b, since a+b+2a+3b-a=x reduces to 2a+4b, where a and b are vectors?

5. Nov 9, 2009

### lanedance

??? a & b are arbitrary scalars - you need to pick vectors from R^5... use a & b to determine them

i'm totally not sure what you mean, but i wouldn't change notation like that mid problem unless you want to make mistakes

6. Nov 9, 2009

### KaiserBrandon

k, ya I didn't think that was right lol. K well wouldn't the set of spanning vectors just be any 5 vectors that satisfy (a,b,2a,3b,-a), like t(u,v,2u,3v,-u), where t is any scalar?

7. Nov 9, 2009

### lanedance

don't just guess, ask yourself why... & why would you need 5 vectors....?

have a look at the questions in post #3, the answer is pretty much determined by the fact any vector in your subspace, can be determined by 2 parameters...

see what vectors you get when you set:
a=1, b=0
a=0, b=1

8. Nov 9, 2009

### KaiserBrandon

alright, so setting a=1, b=0 I get (1,0,2,0,-1), and a=0, b=1, I get (0,1,0,3,0). I need 5 vectors since there are 5 dimensions. I still don't understand what exactly I am trying to find. The only thing I can think of is if you set the vectors a=(1,0,2,0,-1) and b=(0,1,0,3,0), you can have a vector equation a+b+2a+3b-a=w, so then the three vectors would be v1=(1,0,2,0,-1), v2=(0,1,0,3,0), v3=(2,0,4,-2), v4=(0,3,0,9,0), v5=(-1,0,-2,0,1).

9. Nov 9, 2009

### KaiserBrandon

then t(v1+v2+v3+v4+v5) where t is any real number would give a vector which satisfies the parameter x=(a,b,2a,3b,-a)

10. Nov 9, 2009

### lanedance

so you have found two vectors in W:
(1,0,2,0,-1)
(0,1,0,3,0)

What is the definition of linear indpendence?

Can every vector in W be written as a linear combination of these 2 vectors?
- if so, those 2 vectors span W
- if not, you must try & find more linearly independent vectors in W...

why do you need 5 vectors? 5 linearly independent vector will span all of $\mathbb{R}^5$

W is a subspace of $\mathbb{R}^5$, so its dimension must be $\leq$ 5.

11. Nov 9, 2009

### KaiserBrandon

Ok this is making sense now. if I have a linear combination: c1v1+c2v2, where v1 and v2 are those two vectors and c1 and c2 are any real numbers, it gives me a vector which satisfies the conditions of W. So those two vectors do indeed span W. The vectors I gave in post#8 didn't satisfy W for any linear combination. That just leaves the last part of the question asking what type of geometric object W is. I'm unsure of what is meant by "geometric object". Does it mean a line, a plane, the zero vector, or something else?

12. Nov 9, 2009

### lanedance

what do you mean by this? try writing any vector in W as a linear combination of the 2 vectors you found (with unknown multipliers) & see if you can solve it...

what do you think it is? you probably need to work out what the dimension is first