Linear Subspace in R^5 and spanning vectors

In summary, the conversation revolved around finding a linear subspace W of R^5, with vectors of the form x = (a, b, 2a, 3b, -a). It was determined that W meets the criteria for a linear subspace and two spanning vectors, (1,0,2,0,-1) and (0,1,0,3,0), were found. It was also established that W is of dimension 2 and is therefore a plane in \mathbb{R}^5.
  • #1
KaiserBrandon
54
0

Homework Statement


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?


Homework Equations



none

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

This is kind of a hint.
 
  • #3
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)
 
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  • #4
alright, I'm just going to 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
? 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
k, you 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
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
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
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
KaiserBrandon said:
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).
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...


KaiserBrandon said:
I need 5 vectors since there are 5 dimensions. I still don't understand what exactly I am trying to find.

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

W is a subspace of [itex] \mathbb{R}^5[/itex], so its dimension must be [itex] \leq[/itex] 5.
 
  • #11
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
KaiserBrandon said:
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.
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...

KaiserBrandon said:
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?

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

1. What is a linear subspace in R^5?

A linear subspace in R^5 is a subset of the vector space R^5 that satisfies the properties of a vector space. This means that it contains the zero vector, is closed under vector addition and scalar multiplication, and is closed under linear combinations.

2. How is a linear subspace in R^5 different from a vector in R^5?

A vector in R^5 is a single element of the vector space, while a linear subspace in R^5 is a subset of the vector space that contains multiple vectors. A linear subspace must also satisfy the properties of a vector space, while a single vector does not necessarily have to.

3. What are spanning vectors in a linear subspace in R^5?

Spanning vectors are a set of vectors in a linear subspace in R^5 that can be used to create any other vector in the subspace through linear combinations. This means that any vector in the subspace can be written as a sum of scalar multiples of the spanning vectors.

4. How do you determine if a set of vectors span a linear subspace in R^5?

To determine if a set of vectors span a linear subspace in R^5, you can use the spanning set criterion. This means that you must check if every vector in the subspace can be written as a linear combination of the given set of vectors. If this is true, then the set of vectors span the subspace.

5. Can a linear subspace in R^5 have an infinite number of spanning vectors?

Yes, a linear subspace in R^5 can have an infinite number of spanning vectors. This is because a linear subspace can contain an infinite number of vectors, and any set of vectors that can create linear combinations of all of these vectors can be considered a spanning set for the subspace.

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