Register to reply

Determine whether or not something is a subspace

by kesun
Tags: linear algebra, subspace
Share this thread:
kesun
#1
Mar7-09, 03:23 PM
P: 37
My understanding of the subspace still isn't solid enough, so I want to know what I know so far is at least correct.

By definition, a set of vectors S of Rn is called a subspace of Rn iff for all vectors (I will call them x):
1) (x+y) [tex]\in[/tex] S and
2) kx [tex]\in[/tex] S.

Also, the solution set of a homogeneous system is always a subspace.

When I encounter problems such as determine whether or not {(x1,x2,x3)|x1x2=0} is a subspace of its corresponding Rn, I would approach this problem as such:

Since the set has only three vectors, then it's in R3, first of all; then I check for 1): Suppose a set of vectors Y {(y1,y2,y3)|y1y2=0}, then (S+Y)=x1x2+y1y2=0+0=0; for 2): kS=(kx1)(kx2)=(k0)(k0)=0. Therefore it is a subspace of R3.

Is my way of solving this problem correct?
Phys.Org News Partner Science news on Phys.org
Fungus deadly to AIDS patients found to grow on trees
Canola genome sequence reveals evolutionary 'love triangle'
Scientists uncover clues to role of magnetism in iron-based superconductors
ThirstyDog
#2
Mar7-09, 05:31 PM
P: 34
You are right in that you must check your the two conditions. But you must do it for arbitrary vectors, you didn't seem to do this correctly. Consider

[tex] (1,0,0), (0,1,0) \in \{(x_{1},x_{2},x_{3})|x_{1}x_{2}=0\}. [/tex]

If you add them you obtain [tex] (1,1,0) [/tex] which clearly does not have [tex] x_{1}x_{2} = 0 [/tex]. Also the way you present your vectors doesn't seem standard.
de_brook
#3
Mar7-09, 05:35 PM
P: 74
Quote Quote by kesun View Post
My understanding of the subspace still isn't solid enough, so I want to know what I know so far is at least correct.

By definition, a set of vectors S of Rn is called a subspace of Rn iff for all vectors (I will call them x):
1) (x+y) [tex]\in[/tex] S and
2) kx [tex]\in[/tex] S.

Also, the solution set of a homogeneous system is always a subspace.

When I encounter problems such as determine whether or not {(x1,x2,x3)|x1x2=0} is a subspace of its corresponding Rn, I would approach this problem as such:

Since the set has only three vectors, then it's in R3, first of all; then I check for 1): Suppose a set of vectors Y {(y1,y2,y3)|y1y2=0}, then (S+Y)=x1x2+y1y2=0+0=0; for 2): kS=(kx1)(kx2)=(k0)(k0)=0. Therefore it is a subspace of R3.

Is my way of solving this problem correct?
That was a nice attempt but your steps were wrong.
{(x1,x2,x3)|x1x2=0} is a subset of [tex]\mathbb{R}^3\[tex] which satisfies the property that x1x2 = 0. but since x1, x2 are in [tex]\mathbb{R}[tex], then either x1=0 or x2=0 or both equal zero.
To proof that it is a subspace, let y and s be vectors in {(x1,x2,x3)|x1x2=0} such that y = (y1,y2,y3) and s = (s1,s2,s3), then
y+s = (y1+s1,y2+s2,y3+s3)
check (y1+s1)(y2+s2) = y1y2+y1s2+s1y2+s1s2= y1s2+s1y2 (since y1y2=0 and s1s2=0
clear in general; y1s2+s1y2 is not equal to zero. this implies that it does not satisfy the property x1x2 = 0. Hence the set {(x1,x2,x3)|x1x2=0} is not a subspace of [tex]\mathbb{R}^3\[tex]. we don't need to proof the second property

de_brook
#4
Mar7-09, 05:39 PM
P: 74
Determine whether or not something is a subspace

Quote Quote by kesun View Post
My understanding of the subspace still isn't solid enough, so I want to know what I know so far is at least correct.

By definition, a set of vectors S of Rn is called a subspace of Rn iff for all vectors (I will call them x):
1) (x+y) [tex]\in[/tex] S and
2) kx [tex]\in[/tex] S.

Also, the solution set of a homogeneous system is always a subspace.

When I encounter problems such as determine whether or not {(x1,x2,x3)|x1x2=0} is a subspace of its corresponding Rn, I would approach this problem as such:

Since the set has only three vectors, then it's in R3, first of all; then I check for 1): Suppose a set of vectors Y {(y1,y2,y3)|y1y2=0}, then (S+Y)=x1x2+y1y2=0+0=0; for 2): kS=(kx1)(kx2)=(k0)(k0)=0. Therefore it is a subspace of R3.

Is my way of solving this problem correct?
That was a nice attempt but your steps were wrong.
{(x1,x2,x3)|x1x2=0} is a subset of R3 which satisfies the property that x1x2 = 0. but since x1, x2 are in R3 then either x1=0 or x2=0 or both equal zero.
To proof that it is a subspace, let y and s be vectors in {(x1,x2,x3)|x1x2=0} such that y = (y1,y2,y3) and s = (s1,s2,s3), then
y+s = (y1+s1,y2+s2,y3+s3)
check (y1+s1)(y2+s2) = y1y2+y1s2+s1y2+s1s2= y1s2+s1y2 (since y1y2=0 and s1s2=0
clear in general; y1s2+s1y2 is not equal to zero. this implies that it does not satisfy the property x1x2 = 0. Hence the set {(x1,x2,x3)|x1x2=0} is not a subspace of R3. we don't need to proof the second property


Register to reply

Related Discussions
Why R2 is not a subspace of R3? Calculus & Beyond Homework 6
How do i determine if U is a subspace of R3 Precalculus Mathematics Homework 8
What is a subspace of R3? Calculus & Beyond Homework 6
Determine whether this is a subspace Calculus & Beyond Homework 3
Is W a subspace of the vector space? Calculus & Beyond Homework 4