What Determines the Dimension of Subspace S in R^4?

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Homework Help Overview

The problem involves finding a basis for a subspace defined by a set of vectors in R^4, specifically S = {(a+2b,b,-a+b,a+3b) | a,b ∈ R}. The original poster seeks to determine the dimension of this subspace.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to express the subspace S as a span of two vectors, (1,0,-1,1) and (2,1,1,3), and questions whether their approach is correct given the lack of a reference answer.
  • Some participants question the definition of the set S, noting that the initial representation appeared to have only three coordinates, which would not be appropriate for R^4.
  • Others suggest verifying the linear independence of the vectors to confirm they form a basis for the subspace.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the vector representation and the conditions for forming a basis. Clarifications about the dimensionality of the vectors and their independence are being examined, but no consensus has been reached yet.

Contextual Notes

There is a noted confusion regarding the correct representation of the vectors in R^4, which has led to questions about the initial setup of the problem. The original poster has acknowledged a potential error in the notation of the set S.

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Homework Statement



Find a basis for the subspace

S = {(a+2b,b-a+b,a+3b) | a,b [tex]\in[/tex] R } [tex]\subseteq[/tex] R^4

What is the dimension of S?

Homework Equations





The Attempt at a Solution



a(1,0,-1,1) + b(2,1,1,3) , a,b [tex]\in[/tex] R

span { (1,0,-1,1) , (2,1,1,3) }

So I put (1,0,-1,1) as V1 and (2,1,1,3) as V2 and then formed a matrix with V1 and V2 in columns. Then i reduced it to row echelon form. The column corresponding to the leading entry form a basis which is essentially just (1,0,-1,1) and (2,1,1,3) . Thus the dimension is just 2. Am I doing the right thing because I don't have an answer to refer to.
 
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There has to be an error in your definition for this set.
S = {(a+2b,b-a+b,a+3b) | a,b R } R^4

Your vector has only three coordinates, which is too few for it to be in R^4. Are you missing a comma in there?
 
Hi Jeffrey,
jeffreylze said:

Homework Statement



Find a basis for the subspace

S = {(a+2b,b,[/color]-a+b,a+3b) | a,b [tex]\in[/tex] R } [tex]\subseteq[/tex] R^4

What is the dimension of S?

The Attempt at a Solution



a(1,0,-1,1) + b(2,1,1,3) , a,b [tex]\in[/tex] R

span { (1,0,-1,1) , (2,1,1,3) }

So I put (1,0,-1,1) as V1 and (2,1,1,3) as V2 and then formed a matrix with V1 and V2 in columns. Then i reduced it to row echelon form. The column corresponding to the leading entry form a basis which is essentially just (1,0,-1,1) and (2,1,1,3) . Thus the dimension is just 2. Am I doing the right thing because I don't have an answer to refer to.
The first step you have done is shown that S = span {(1,0,-1,1), (2,1,1,3)}. Then if (1,0,-1,1) and (2,1,1,3) are linearly independent, they form a basis (recall the definition) for S.

It is correct that n vectors are linearly independent if and only if the matrix formed with them as its columns has rank n, but in this case it is probably simpler just to recognise that two non-zero vectors are linearly independent iff they're not scalar multiples of each other.
 
Mark44 said:
There has to be an error in your definition for this set.


Your vector has only three coordinates, which is too few for it to be in R^4. Are you missing a comma in there?

Oops, my bad. Yes, there is a comma in there. Should be, S = {(a+2b,b,-a+b,a+3b) | a,b R } R^4
 

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