New Reply

Linear Algebra: Basis and Dimension problem

 
Share Thread Thread Tools
Mar3-08, 03:15 PM   #1
 

Linear Algebra: Basis and Dimension problem


1. The problem statement, all variables and given/known data

Given the vectors: x1=(3,-2,4), x2=(-3,2,-4) and x3=(-6,4,-8) , what is the dimension of Span(x1,x2,x3)

2. Relevant equations



3. The attempt at a solution

I know x1,x2 and x3 are Linearly dependent since its determinant is zero. There are a total of 3 vectors in the spanning set. I thought the number of dimensions would be 3. But my the back of my book says the number of dimensions is one.
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> Galaxies fed by funnels of fuel
>> The better to see you with: Scientists build record-setting metamaterial flat lens
>> Google eyes emerging markets networks
Mar3-08, 03:43 PM   #2
 
x1, x2 and x3 are scalar multiples of each other (check this for yourself). Therefore, the dimension is 1.
Mar3-08, 03:50 PM   #3
 
Quote by jhicks View Post
x1, x2 and x3 are scalar multiples of each other (check this for yourself). Therefore, the dimension is 1.
I do not understand how x1 , x2 and x3 being scalar multiples of each other makes the number of dimensions 1.
Mar3-08, 04:19 PM   #4
 

Linear Algebra: Basis and Dimension problem


basically, x1, x2 and x3 are linearly dependent in that each can be written in terms of only one other multiplied by a scaling constant. Since the dimension of the span is how many linearly independent vectors there are (only one in this case), the dimension of the span is 1.
Mar3-08, 04:28 PM   #5
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
The definition of dimension of a space is the number of vectors in a basis. A basis is any set of vectors that both spans the space and is independent. The given set is NOT a basis specifically because it in not independent- as you say, it is dependent. If a set of vectors is dependent, at least one of the vectors can be written as a linear combination of the others and can be dropped from the set. In this case, (-3, 2, -4) = (-1)(3, -2, 4) so any vector that can be written as a linear combination of the three given vectors can be written without (-3, 2, -4): {(3, -2, 4), (-6, 4, 8)} also spans the space. But that set is also dependent: (-6, 4, 8)= -2(3, -2, 4) so any vector in its span can be written as a multiple of (3, -2, 4). That means that the set {(3, -2, 4)} both spans the space and is trivially independent. That set is a basis for the space and, because it contains only one vector, the dimension of the space is 1.
Nov8-10, 02:26 AM   #6
 
you said that there are two conditions in order to be a a basis one is that the set should be linearly independent and the other condition it should span the space, my question is what is the space these vectors given do they span ? why you did not concentrate on this part of the definition of basis and only on linear independence? and these vectors are in R3, why the dimension is not 3 for example ?
New Reply
Thread Tools


Similar Threads for: Linear Algebra: Basis and Dimension problem
Thread Forum Replies
Linear Algebra Subspace Basis Problem Calculus & Beyond Homework 5
Dimension - Linear Algebra Calculus & Beyond Homework 2
Linear Algebra - Basis Calculus & Beyond Homework 9
Linear Algebra -- Basis Introductory Physics Homework 3
Linear Algebra -- Dimension, Basis Introductory Physics Homework 3