Linear Subspace in R^5 and spanning vectors

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Homework Help Overview

The problem involves determining whether a set of vectors defined in the form x = (a, b, 2a, 3b, -a) constitutes a linear subspace of R^5. Participants are also tasked with identifying spanning vectors for this set and exploring the geometric nature of the subspace.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the criteria for a linear subspace and the definition of spanning vectors. There are attempts to express vectors in terms of parameters a and b, with some questioning the need for a specific number of spanning vectors. Others suggest setting values for a and b to derive specific vectors in the subspace.

Discussion Status

The discussion is active, with participants exploring different approaches to identify spanning vectors. Some have proposed specific vectors derived from parameter values, while others are questioning the dimensionality of the subspace and the implications for the number of spanning vectors needed. There is no explicit consensus yet on the final characterization of the subspace.

Contextual Notes

Participants are navigating the definitions and properties of linear combinations and independence, as well as the geometric interpretation of the subspace, which remains under discussion. The constraints of the problem include the requirement to find spanning vectors and to understand the geometric object represented by W.

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

This is kind of a hint.
 
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:
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?
 
? 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
 
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?
 
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
 
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).
 
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
 

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